Develop Reference

Dijkstra adjacency list

I have run into a problem converting pseudocode of Dijkstras algorithm into actual code. I was given and adjacency list such as "Location - adjacent location - distance to location," example for one node: AAA AAC 180 AAD 242 AAH 40.
My task was to read a file organized as adjacency list as described, and compute the shortest path from one node to another.
Here is the Dijkstra pseudocode:
void dijkstra( Vertex s )
{
for each Vertex v
{
v.dist = INFINITY;
v.known = false;
}
s.dist = 0;
while( there is an unknown distance vertex )
{
Vertex v = smallest unknown distance vertex;
v.known = true;
for each Vertex w adjacent to v
if( !w.known )
{
DistType cvw = cost of edge from v to w;
if( v.dist + cvw < w.dist )
{
// Update w
decrease( w.dist to v.dist + cvw );
w.path = v;
}
}
}
}
im having the most trouble with the line "for each Vertex w adjacent to v"
Here is my nonworking code:
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.util.LinkedList;
import java.util.List;
import java.util.ListIterator;
public class Dijkstra {
public static boolean isInteger(String s) {
return isInteger(s, 10);
}
public static boolean isInteger(String s, int radix) {
if (s.isEmpty())
return false;
for (int i = 0; i < s.length(); i++) {
if (i == 0 && s.charAt(i) == '-') {
if (s.length() == 1)
return false;
else
continue;
}
if (Character.digit(s.charAt(i), radix) < 0)
return false;
}
return true;
}
public static void dijkstra(Vertex[] a, Vertex s, int lineCount) {
int i = 0;
while (i < (lineCount)) // each Vertex v
{
a[i].dist = Integer.MAX_VALUE;
a[i].known = false;
i++;
}
s.dist = 0;
int min = Integer.MAX_VALUE; //
while (!(a[0].known == true && a[1].known == true && a[2].known == true && a[3].known == true
&& a[4].known == true && a[5].known == true && a[6].known == true && a[7].known == true
&& a[8].known == true && a[9].known == true && a[10].known == true && a[11].known == true
&& a[12].known == true)) {
System.out.println("here");
for (int b = 0; b < lineCount; b++) {
if (a[b].dist < min && a[b].known == false) {
min = a[b].dist;
}
}
int c = 0;
while (c < lineCount) {
if (a[c].dist == min && a[c].known == false) {
break;
}
c++;
}
System.out.println(min);
a[c].known = true;
int adjSize = a[c].adj.size();
int current = 0;
System.out.println(adjSize);
while (current < adjSize - 1) {
String currentAdjacent = (String) a[c].adj.get(current);
int p = 0;
while (p < lineCount) {
if (a[p].name.equals(currentAdjacent)) {
if (!a[p].known) {
String cvwString = (String) a[c].distance.get(current);
int cvw = Integer.parseInt(cvwString);
System.out.println(" This is cvw" + cvw);
System.out.println("Here2");
if (a[c].dist + cvw < a[p].dist) {
a[p].dist = a[c].dist + cvw;
a[p].path = a[c];
}
}
}
p++;
}
current++;
}
}
}
public static class Vertex {
public List adj; // Adjacency list
public List distance;
public boolean known;
public int dist; // DistType is probably int
public Vertex path;
public String name;
// Other fields and methods as needed
}
public static void printPath(Vertex v) {
if (v.path != null) {
printPath(v.path);
System.out.print(" to ");
}
System.out.print(v);
}
public static void main(String[] args) throws IOException {
int lineCounter = 0;
BufferedReader br = new BufferedReader(new FileReader("airport.txt"));
try {
StringBuilder sb = new StringBuilder();
String line = br.readLine();
while (line != null) {
sb.append(line);
sb.append(System.lineSeparator());
line = br.readLine();
lineCounter = lineCounter + 1;
}
Vertex[] arr = new Vertex[lineCounter];
for (int i = 0; i < lineCounter; i++) {
arr[i] = new Vertex();
arr[i].adj = new LinkedList<String>();
arr[i].distance = new LinkedList<Integer>();
}
;
//
int arrayCounter = 0;
String everything = sb.toString();
String[] lines = everything.split("\\s*\\r?\\n\\s*");
for (String line1 : lines) {
arr[arrayCounter] = new Vertex();
arr[arrayCounter].adj = new LinkedList<String>();
arr[arrayCounter].distance = new LinkedList<Integer>();
String[] result = line1.split("\\s+");
for (int x = 0; x < result.length; x++) {
if (x == 0) {
arr[arrayCounter].name = result[0];
continue;
} else if (isInteger(result[x])) {
arr[arrayCounter].distance.add(result[x]);
continue;
} else {
arr[arrayCounter].adj.add(result[x]);
continue;
}
}
arrayCounter++;
}
for (int i = 0; i < 12; i++) {
System.out.println(arr[i].name);
}
System.out.println(lineCounter);
dijkstra(arr, arr[3], lineCounter - 1);
printPath(arr[11]);
} finally {
br.close();
}
}
}
Using my vertex class as is I was using a series of while loops to first, traverse the adjacency strings stored in a linked list while comparing to see which vertex is equivalent to the adjacency list string. Is there a better way to code "for each Vertex w adjacent to v" using my Vertex class? And apologies ahead for messy code and any others style sins i may have committed. Thanks!

To solve this problem you need a bunch of "Node" objects, stored in a HashMap, keyed on Source Location.
In the node, you need a collection of references to adjacent "Node" objects (or at least their "key" so you can write logic against it. The "Node" also needs to know it's location and distance to each "adjacent" node. Think Lundon Underground Tube Maps - each station connects to at least one other station. Usually two or more. Therefore, adjacent nodes to tube stations are the immediate next stops you can get to from that station.
Once you have that data structure in place, you can then use a recursive routine to iterate through each individual node. It should then iterate through each child node (aka adjacent node), and track distances from the initial (source) node to the current node by storing this data in a HashMap and using the current accumulated distance whilst recursing (or "walking" the graph"). This tracking information should be part of your method signature when recursing. You will also need to track the current path you have taken when recursing, in order to avoid circular loops (which will ultimately and ironically cause a StackOverflowError). You can do this by using a HashSet. This Set should track the source and current node's location as the entry key. If you see this present during your recursion, then you have already seen it, so don't continue processing.
I'm not going to code the solution for you because I suspect that you ask more specific questions as you work your way through understanding the answer, which are very likely answered elsewhere.

Insert a new string to the trie graph

I am trying to implement the insert method of the Patricia Trie data structure and I am trying to handle this case:
first string: abaxyzalexsky,
second string: abaxyzalex,
third string: abaxyz,
fourth string: aba
I want to mark the trie as the following aba-xyz-alex-sky after inserting the fourth string, but I don't know how can I get it work.
How can I mark the words in the trie in the case above?
public void insert(String s) {
if (nodeRoot == null) {
nodeRoot = new TrieNode(s);
nodeRoot.isWord = true;
} else {
insert(nodeRoot, s);
}
}
private void insert(TrieNode node, String s) {
int len1 = node.edge.length();
int len2 = s.length();
int len = Math.min(len1, len2);
ArrayList<TrieNode> nextNode = node.getNext();
for (int index = 0; index < len; index++) {
if (s.charAt(index) != node.edge.charAt(index)) {
// In case the both words have common substrings and after the
// common substrings the words are split. For example abad, abac
} else if (index == (s.length() - 1)
|| index == (node.edge.length() - 1)) {
// In case the node just needs one path since one word is
// substring of the other.
// For example (aba and abac)
if (len1 > len2) {
// node edge string is longer than the inserted one. For example (abac
// and aba).
String samesubString = node.edge.substring(0, index + 1);
String different = node.edge.substring(index + 1);
node.edge = samesubString;
if (node.getNext() != null && !node.getNext().isEmpty()) {
for (TrieNode subword : node.getNext()) {
//I am here when I insert the third string. The code below retrives wrong data structure.
TrieNode node1 = new TrieNode(different);
node1.isWord = true;
node1.next.add(subword);
node.next.add(node1);
}
} else {
TrieNode leaf = new TrieNode(different);
leaf.isWord = true;
node.next.add(leaf);
for (TrieNode subword : node.getNext()) {
System.out.println(node.getEdge() + "---"
+ subword.getEdge());
}
}
} else {
// new inserted string value is longer. For example (aba
// and abac).
}
} else {
System.out.println("The strings are the same - " + index);
}
}
}
NodeTrie class
package patriciaTrie;
import java.util.ArrayList;
public class TrieNode {
ArrayList<TrieNode> next = new ArrayList<TrieNode>();
String edge;
boolean isWord;
TrieNode(String edge){
this.edge = edge;
}
public ArrayList<TrieNode> getNext() {
return next;
}
public void setNext(ArrayList<TrieNode> next) {
this.next = next;
}
public String getEdge() {
return edge;
}
public void setEdge(String edge) {
this.edge = edge;
}
}

insert method of the Trie data structure

I am trying ti implement the insert method of the Patricia Trie data structure but I have the feeling I wrote to many code lines. Please can someone tell me where can I call the method insert(TrieNode nodeRoot, String s) rekursiv?
Code:
private void insert(TrieNode nodeRoot, String s) {
int len1 = nodeRoot.value.length();
int len2 = s.length();
int len = Math.min(len1, len2);
for (int index = 0; index < len; index++) {
if (s.charAt(index) != nodeRoot.value.charAt(index)) {
// In case the both words have common substrings and after the
// common substrings the words are split.
String samesubString = s.substring(0, index);
String substringSplit1 = nodeRoot.value.substring(index);
String substringSplit2 = s.substring(index);
if (!samesubString.isEmpty()) {
nodeRoot.value = samesubString;
}
TrieNode nodeLeft = new TrieNode(substringSplit1);
nodeLeft.isWord = true;
TrieNode nodeRight = new TrieNode(substringSplit2);
nodeRight.isWord = true;
if (nodeRoot.getNext() != null && !nodeRoot.getNext().isEmpty()) {
checkTheValieAvialable(nodeRoot, s, nodeRight);
} else {
nodeRoot.next.add(nodeLeft);
nodeRoot.next.add(nodeRight);
for (TrieNode subword : nodeRoot.getNext()) {
System.out.println(nodeRoot.getValue() + "---"
+ subword.getValue());
}
}
break;
} else if (index == (s.length() - 1)
|| index == (nodeRoot.value.length() - 1)) {
// In case the node just needs one path since one word is
// substring of the other.
// For example (aba and abac)
if (len1 > len2) {
// root value is longer
System.out.println("root value is longer");
String samesubString = nodeRoot.value.substring(0,
index + 1);
String different = nodeRoot.value.substring(index + 1);
if (nodeRoot.getNext() != null
&& !nodeRoot.getNext().isEmpty()) {
for (TrieNode subword : nodeRoot.getNext()) {
String subword2 = subword.getValue();
boolean contains = different.contains(subword2);
if (contains) {
String[] split = different.split(subword2);
TrieNode leaf1 = new TrieNode(split[1]);
leaf1.isWord = true;
subword.next.add(leaf1);
System.out.println("Test.");
}
}
} else {
String substringSplit1 = nodeRoot.value.substring(index + 1);
nodeRoot.value = samesubString;
TrieNode leaf = new TrieNode(substringSplit1);
leaf.isWord = true;
nodeRoot.next.add(leaf);
for (TrieNode subword : nodeRoot.getNext()) {
System.out.println(nodeRoot.getValue() + "---"
+ subword.getValue());
}
}
String substringSplit1 = nodeRoot.value
.substring(index + 1);
nodeRoot.value = samesubString;
nodeRoot.isWord = true;
TrieNode leaf = new TrieNode(substringSplit1);
leaf.isWord = true;
nodeRoot.next.add(leaf);
for (TrieNode subword : nodeRoot.getNext()) {
System.out.println(nodeRoot.getValue() + "---"
+ subword.getValue());
}
} else {
// new inserted string value is longer. For example (abac and aba).
System.out.println("instered is longer");
String samesubString = s.substring(0, index + 1);
String different = s.substring(index + 1);
if (nodeRoot.getNext() != null
&& !nodeRoot.getNext().isEmpty()) {
for (TrieNode subword : nodeRoot.getNext()) {
String subword2 = subword.getValue();
boolean contains = different.contains(subword2);
if (contains) {
String[] split = different.split(subword2);
TrieNode leaf1 = new TrieNode(split[1]);
leaf1.isWord = true;
subword.next.add(leaf1);
System.out.println("Test.");
}
}
} else {
String substringSplit1 = s.substring(index + 1);
s = samesubString;
TrieNode parentLeaf = new TrieNode(s);
parentLeaf.isWord = true;
TrieNode leaf = new TrieNode(substringSplit1);
leaf.isWord = true;
nodeRoot.next.add(leaf);
for (TrieNode subword : nodeRoot.getNext()) {
System.out.println(nodeRoot.getValue() + "---"
+ subword.getValue());
}
}
}
} else {
System.out.println("They are the same - " + index);
}
}
}
TrieNode class:
package patriciaTrie;
import java.util.ArrayList;
public class TrieNode {
ArrayList<TrieNode> next = new ArrayList<TrieNode>();
String value;
boolean isWord;
TrieNode(String value){
this.value = value;
}
public ArrayList<TrieNode> getNext() {
return next;
}
public void setNext(ArrayList<TrieNode> next) {
this.next = next;
}
public String getValue() {
return value;
}
public void setValue(String value) {
this.value = value;
}
}

While using recursion please consider the steps:
Base condition
Logic (if any)
Recursive call.
Ex. for factorial of number:
int fact(int n)
{
if(n==0 || n==1)
return 1; // Base condition
return n * fact(n-1); // Recursive call
}
Applying the same concept in Trie:
base condition is: while traversing through a path, if we have reached leaf, current string is not in trie, then create a new edge or node and add remaining character to it.
Recursively call the insert if we have found a matching node. And if a matching node doen't exist create a new path with common parent.
You can take help from link : http://www.geeksforgeeks.org/trie-insert-and-search/
The best way to approach to problem recursively is to identify base condition in a problem.

Java Binary Tree entered in a specific order

I am trying to complete an assignment where I need to write a Java program to take a string from the command line, and implement it as a Binary Tree in a specific order, then get the depth of the binary tree.
For example: "((3(4))7((5)9))"
would be entered as a tree with 7 as the root, 3 and 9 as the children, and 4 as a right child of 3, and 5 as a left child of 9.
My code is below.. The problem I am having is that, because I am basing my checks off of finding a right bracket, I am unsure how to get the elements correctly when they are not directly preceding the brackets, such as the 3 in the above string. Any direction would be greatly appreciated..
class Node {
int value;
Node left, right;
}
class BST {
public Node root;
// Add Node to Tree
public void add(int n) {
if (root == null) {
root = new Node( );
root.value = n;
}
else {
Node marker = root;
while (true) {
if (n < marker.value) {
if (marker.left == null) {
marker.left = new Node( );
marker.left.value = n;
break;
} else {
marker = marker.left;
}
} else {
if (marker.right == null) {
marker.right = new Node( );
marker.right.value = n;
break;
} else {
marker = marker.right;
}
}
}
}
} // End ADD
//Find Height of Tree
public int height(Node t) {
if (t.left == null && t.right == null) return 0;
if (t.left == null) return 1 + height(t.right);
if (t.right == null) return 1 + height(t.left);
return 1 + Math.max(height(t.left), height(t.right));
} // End HEIGHT
// Check if string contains an integer
public static boolean isInt(String s) {
try {
Integer.parseInt(s);
}
catch(NumberFormatException e) {
return false;
}
return true;
} // End ISINT
public int elementCount(String[] a) {
int count = 0;
for (int i = 0; i < a.length; i++) {
if (isInt(a[i])) count++;
}
return count;
}
} // End BST Class
public class Depth {
public static void main(String[] args) {
String[] a = args[0].split(" ");
BST tree = new BST();
int[] bcount = new int[10];
int[] elements = new int[10];
int x = 0, bracketcount = 0;
// Display entered string
System.out.print("Entered Format: ");
for (int j=0; j < a.length; j++) {
System.out.print(a[j]);
}
for (int i=0; i < a.length; i++) {
char c = a[i].charAt(0);
switch (c)
{
case '(':
bracketcount++;
break;
case ')':
if (isInt(a[i-1])) {
bcount[x] = bracketcount--;
elements[x++] = Integer.parseInt(a[i-1]);
}
break;
case '1':
case '7':
default : // Illegal character
if ( (a[i-1].charAt(0) == ')') && (a[i+1].charAt(0) == '(') ) {
bcount[x] = bracketcount;
elements[x++] = Integer.parseInt(a[i]);
}
break;
}
}
System.out.println("\nTotal elements: " + tree.elementCount(a));
// Display BracketCounts
for (int w = 0; w < x; w++) {
System.out.print(bcount[w] + " ");
}
System.out.println(" ");
// Display Elements Array
for (int w = 0; w < x; w++) {
System.out.print(elements[w] + " ");
}
System.out.println("\nDepth: " + tree.height(tree.root));
// Build the tree
for (int y = 0; y < x-1; y++) {
for (int z = 1; z < tree.height(tree.root); z++) {
if (bcount[y] == z) {
tree.add(elements[y]);
}
}
}
} // End Main Function
public static boolean isInt(String s) {
try {
Integer.parseInt(s);
}
catch(NumberFormatException e) {
return false;
}
return true;
}
} // End Depth Class

I would do a couple of statements to get access to a tree with that kind of shape:
For input string : input= "((3(4))7((5)9))"
You could do :
public class Trial {
/**
* #param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
String input = "((3(4))7((5)9))";
String easier = input.replaceAll("\\(\\(", "");
String evenEasier = easier.replaceAll("\\)\\)", "");
System.out.println(evenEasier);
int firstVal = Integer.parseInt(evenEasier.substring(0, 1));
int firstBracketVal = Integer.parseInt(evenEasier.substring(2, 3));
int middleVal = Integer.parseInt(evenEasier.substring(3, 4));
int secondBracketVal = Integer.parseInt(evenEasier.substring(4,5));
int lastVal = Integer.parseInt(evenEasier.substring(6));
System.out.println("First Val:"+firstVal);
System.out.println("First bracket Val:"+firstBracketVal);
System.out.println("Middle Val:"+middleVal);
System.out.println("Second Bracket Val:"+secondBracketVal);
System.out.println("Last Val:"+lastVal);
}
}
This however would only ever work for entries in that specific format, if that were to change, or the length of the input goes up - this would work a bit or break.....If you need to be able to handle more complicated trees as input in this format a bit more thought would be needed on how to best handle and convert into your internal format for processing.

pseudocode:
function getNode(Node)
get one char;
if (the char is "(")
getNode(Node.left);
get one char;
end if;
Node.value = Integer(the char);
get one char;
if (the char is "(")
getNode(Node.right);
get one char;
end if;
//Now the char is ")" and useless.
end function
Before calling this function, you should get a "(" first.
In this method, the framwork of a Node in string is "[leftchild or NULL] value [rightchild or NULL])".
"("is not belong to the Node, but ")" is.

How to print binary tree diagram in Java?

How can I print a binary tree in Java so that the output is like:
4
/ \
2 5
My node:
public class Node<A extends Comparable> {
Node<A> left, right;
A data;
public Node(A data){
this.data = data;
}
}

Print a [large] tree by lines.
output example:
z
├── c
│   ├── a
│   └── b
├── d
├── e
│   └── asdf
└── f
code:
public class TreeNode {
final String name;
final List<TreeNode> children;
public TreeNode(String name, List<TreeNode> children) {
this.name = name;
this.children = children;
}
public String toString() {
StringBuilder buffer = new StringBuilder(50);
print(buffer, "", "");
return buffer.toString();
}
private void print(StringBuilder buffer, String prefix, String childrenPrefix) {
buffer.append(prefix);
buffer.append(name);
buffer.append('\n');
for (Iterator<TreeNode> it = children.iterator(); it.hasNext();) {
TreeNode next = it.next();
if (it.hasNext()) {
next.print(buffer, childrenPrefix + "├── ", childrenPrefix + "│ ");
} else {
next.print(buffer, childrenPrefix + "└── ", childrenPrefix + " ");
}
}
}
}
P.S. This answer doesn't exactly focus on "binary" trees -- instead, it prints all kinds of trees. Solution is inspired by the "tree" command in linux.

I've created simple binary tree printer. You can use and modify it as you want, but it's not optimized anyway. I think that a lot of things can be improved here ;)
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
public class BTreePrinterTest {
private static Node<Integer> test1() {
Node<Integer> root = new Node<Integer>(2);
Node<Integer> n11 = new Node<Integer>(7);
Node<Integer> n12 = new Node<Integer>(5);
Node<Integer> n21 = new Node<Integer>(2);
Node<Integer> n22 = new Node<Integer>(6);
Node<Integer> n23 = new Node<Integer>(3);
Node<Integer> n24 = new Node<Integer>(6);
Node<Integer> n31 = new Node<Integer>(5);
Node<Integer> n32 = new Node<Integer>(8);
Node<Integer> n33 = new Node<Integer>(4);
Node<Integer> n34 = new Node<Integer>(5);
Node<Integer> n35 = new Node<Integer>(8);
Node<Integer> n36 = new Node<Integer>(4);
Node<Integer> n37 = new Node<Integer>(5);
Node<Integer> n38 = new Node<Integer>(8);
root.left = n11;
root.right = n12;
n11.left = n21;
n11.right = n22;
n12.left = n23;
n12.right = n24;
n21.left = n31;
n21.right = n32;
n22.left = n33;
n22.right = n34;
n23.left = n35;
n23.right = n36;
n24.left = n37;
n24.right = n38;
return root;
}
private static Node<Integer> test2() {
Node<Integer> root = new Node<Integer>(2);
Node<Integer> n11 = new Node<Integer>(7);
Node<Integer> n12 = new Node<Integer>(5);
Node<Integer> n21 = new Node<Integer>(2);
Node<Integer> n22 = new Node<Integer>(6);
Node<Integer> n23 = new Node<Integer>(9);
Node<Integer> n31 = new Node<Integer>(5);
Node<Integer> n32 = new Node<Integer>(8);
Node<Integer> n33 = new Node<Integer>(4);
root.left = n11;
root.right = n12;
n11.left = n21;
n11.right = n22;
n12.right = n23;
n22.left = n31;
n22.right = n32;
n23.left = n33;
return root;
}
public static void main(String[] args) {
BTreePrinter.printNode(test1());
BTreePrinter.printNode(test2());
}
}
class Node<T extends Comparable<?>> {
Node<T> left, right;
T data;
public Node(T data) {
this.data = data;
}
}
class BTreePrinter {
public static <T extends Comparable<?>> void printNode(Node<T> root) {
int maxLevel = BTreePrinter.maxLevel(root);
printNodeInternal(Collections.singletonList(root), 1, maxLevel);
}
private static <T extends Comparable<?>> void printNodeInternal(List<Node<T>> nodes, int level, int maxLevel) {
if (nodes.isEmpty() || BTreePrinter.isAllElementsNull(nodes))
return;
int floor = maxLevel - level;
int endgeLines = (int) Math.pow(2, (Math.max(floor - 1, 0)));
int firstSpaces = (int) Math.pow(2, (floor)) - 1;
int betweenSpaces = (int) Math.pow(2, (floor + 1)) - 1;
BTreePrinter.printWhitespaces(firstSpaces);
List<Node<T>> newNodes = new ArrayList<Node<T>>();
for (Node<T> node : nodes) {
if (node != null) {
System.out.print(node.data);
newNodes.add(node.left);
newNodes.add(node.right);
} else {
newNodes.add(null);
newNodes.add(null);
System.out.print(" ");
}
BTreePrinter.printWhitespaces(betweenSpaces);
}
System.out.println("");
for (int i = 1; i <= endgeLines; i++) {
for (int j = 0; j < nodes.size(); j++) {
BTreePrinter.printWhitespaces(firstSpaces - i);
if (nodes.get(j) == null) {
BTreePrinter.printWhitespaces(endgeLines + endgeLines + i + 1);
continue;
}
if (nodes.get(j).left != null)
System.out.print("/");
else
BTreePrinter.printWhitespaces(1);
BTreePrinter.printWhitespaces(i + i - 1);
if (nodes.get(j).right != null)
System.out.print("\\");
else
BTreePrinter.printWhitespaces(1);
BTreePrinter.printWhitespaces(endgeLines + endgeLines - i);
}
System.out.println("");
}
printNodeInternal(newNodes, level + 1, maxLevel);
}
private static void printWhitespaces(int count) {
for (int i = 0; i < count; i++)
System.out.print(" ");
}
private static <T extends Comparable<?>> int maxLevel(Node<T> node) {
if (node == null)
return 0;
return Math.max(BTreePrinter.maxLevel(node.left), BTreePrinter.maxLevel(node.right)) + 1;
}
private static <T> boolean isAllElementsNull(List<T> list) {
for (Object object : list) {
if (object != null)
return false;
}
return true;
}
}
Output 1 :
2
/ \
/ \
/ \
/ \
7 5
/ \ / \
/ \ / \
2 6 3 6
/ \ / \ / \ / \
5 8 4 5 8 4 5 8
Output 2 :
2
/ \
/ \
/ \
/ \
7 5
/ \ \
/ \ \
2 6 9
/ \ /
5 8 4

I've made an improved algorithm for this, which handles nicely nodes with different size. It prints top-down using lines.
package alg;
import java.util.ArrayList;
import java.util.List;
/**
* Binary tree printer
*
* #author MightyPork
*/
public class TreePrinter
{
/** Node that can be printed */
public interface PrintableNode
{
/** Get left child */
PrintableNode getLeft();
/** Get right child */
PrintableNode getRight();
/** Get text to be printed */
String getText();
}
/**
* Print a tree
*
* #param root
* tree root node
*/
public static void print(PrintableNode root)
{
List<List<String>> lines = new ArrayList<List<String>>();
List<PrintableNode> level = new ArrayList<PrintableNode>();
List<PrintableNode> next = new ArrayList<PrintableNode>();
level.add(root);
int nn = 1;
int widest = 0;
while (nn != 0) {
List<String> line = new ArrayList<String>();
nn = 0;
for (PrintableNode n : level) {
if (n == null) {
line.add(null);
next.add(null);
next.add(null);
} else {
String aa = n.getText();
line.add(aa);
if (aa.length() > widest) widest = aa.length();
next.add(n.getLeft());
next.add(n.getRight());
if (n.getLeft() != null) nn++;
if (n.getRight() != null) nn++;
}
}
if (widest % 2 == 1) widest++;
lines.add(line);
List<PrintableNode> tmp = level;
level = next;
next = tmp;
next.clear();
}
int perpiece = lines.get(lines.size() - 1).size() * (widest + 4);
for (int i = 0; i < lines.size(); i++) {
List<String> line = lines.get(i);
int hpw = (int) Math.floor(perpiece / 2f) - 1;
if (i > 0) {
for (int j = 0; j < line.size(); j++) {
// split node
char c = ' ';
if (j % 2 == 1) {
if (line.get(j - 1) != null) {
c = (line.get(j) != null) ? '┴' : '┘';
} else {
if (j < line.size() && line.get(j) != null) c = '└';
}
}
System.out.print(c);
// lines and spaces
if (line.get(j) == null) {
for (int k = 0; k < perpiece - 1; k++) {
System.out.print(" ");
}
} else {
for (int k = 0; k < hpw; k++) {
System.out.print(j % 2 == 0 ? " " : "─");
}
System.out.print(j % 2 == 0 ? "┌" : "┐");
for (int k = 0; k < hpw; k++) {
System.out.print(j % 2 == 0 ? "─" : " ");
}
}
}
System.out.println();
}
// print line of numbers
for (int j = 0; j < line.size(); j++) {
String f = line.get(j);
if (f == null) f = "";
int gap1 = (int) Math.ceil(perpiece / 2f - f.length() / 2f);
int gap2 = (int) Math.floor(perpiece / 2f - f.length() / 2f);
// a number
for (int k = 0; k < gap1; k++) {
System.out.print(" ");
}
System.out.print(f);
for (int k = 0; k < gap2; k++) {
System.out.print(" ");
}
}
System.out.println();
perpiece /= 2;
}
}
}
To use this for your Tree, let your Node class implement PrintableNode.
Example output:
2952:0
┌───────────────────────┴───────────────────────┐
1249:-1 5866:0
┌───────────┴───────────┐ ┌───────────┴───────────┐
491:-1 1572:0 4786:1 6190:0
┌─────┘ └─────┐ ┌─────┴─────┐
339:0 5717:0 6061:0 6271:0

public static class Node<T extends Comparable<T>> {
T value;
Node<T> left, right;
public void insertToTree(T v) {
if (value == null) {
value = v;
return;
}
if (v.compareTo(value) < 0) {
if (left == null) {
left = new Node<T>();
}
left.insertToTree(v);
} else {
if (right == null) {
right = new Node<T>();
}
right.insertToTree(v);
}
}
public void printTree(OutputStreamWriter out) throws IOException {
if (right != null) {
right.printTree(out, true, "");
}
printNodeValue(out);
if (left != null) {
left.printTree(out, false, "");
}
}
private void printNodeValue(OutputStreamWriter out) throws IOException {
if (value == null) {
out.write("<null>");
} else {
out.write(value.toString());
}
out.write('\n');
}
// use string and not stringbuffer on purpose as we need to change the indent at each recursion
private void printTree(OutputStreamWriter out, boolean isRight, String indent) throws IOException {
if (right != null) {
right.printTree(out, true, indent + (isRight ? " " : " | "));
}
out.write(indent);
if (isRight) {
out.write(" /");
} else {
out.write(" \\");
}
out.write("----- ");
printNodeValue(out);
if (left != null) {
left.printTree(out, false, indent + (isRight ? " | " : " "));
}
}
}
will print:
/----- 20
| \----- 15
/----- 14
| \----- 13
/----- 12
| | /----- 11
| \----- 10
| \----- 9
8
| /----- 7
| /----- 6
| | \----- 5
\----- 4
| /----- 3
\----- 2
\----- 1
for the input
8
4
12
2
6
10
14
1
3
5
7
9
11
13
20
15
this is a variant from #anurag's answer - it was bugging me to see the extra |s

Adapted from Vasya Novikov's answer to make it more binary, and use a StringBuilder for efficiency (concatenating String objects together in Java is generally inefficient).
public StringBuilder toString(StringBuilder prefix, boolean isTail, StringBuilder sb) {
if(right!=null) {
right.toString(new StringBuilder().append(prefix).append(isTail ? "│ " : " "), false, sb);
}
sb.append(prefix).append(isTail ? "└── " : "┌── ").append(value.toString()).append("\n");
if(left!=null) {
left.toString(new StringBuilder().append(prefix).append(isTail ? " " : "│ "), true, sb);
}
return sb;
}
#Override
public String toString() {
return this.toString(new StringBuilder(), true, new StringBuilder()).toString();
}
Output:
│ ┌── 7
│ ┌── 6
│ │ └── 5
└── 4
│ ┌── 3
└── 2
└── 1
└── 0

I found VasyaNovikov's answer very useful for printing a large general tree, and modified it for a binary tree
Code:
class TreeNode {
Integer data = null;
TreeNode left = null;
TreeNode right = null;
TreeNode(Integer data) {this.data = data;}
public void print() {
print("", this, false);
}
public void print(String prefix, TreeNode n, boolean isLeft) {
if (n != null) {
System.out.println (prefix + (isLeft ? "|-- " : "\\-- ") + n.data);
print(prefix + (isLeft ? "| " : " "), n.left, true);
print(prefix + (isLeft ? "| " : " "), n.right, false);
}
}
}
Sample output:
\-- 7
|-- 3
| |-- 1
| | \-- 2
| \-- 5
| |-- 4
| \-- 6
\-- 11
|-- 9
| |-- 8
| \-- 10
\-- 13
|-- 12
\-- 14

michal.kreuzman nice one I will have to say. It was useful.
However, the above works only for single digits: if you are going to use more than one digit, the structure is going to get misplaced since you are using spaces and not tabs.
As for my later codes I needed more digits than only 2, so I made a program myself.
It has some bugs now, again right now I am feeling lazy to correct them but it prints very beautifully and the nodes can take a larger number of digits.
The tree is not going to be as the question mentions but it is 270 degrees rotated :)
public static void printBinaryTree(TreeNode root, int level){
if(root==null)
return;
printBinaryTree(root.right, level+1);
if(level!=0){
for(int i=0;i<level-1;i++)
System.out.print("|\t");
System.out.println("|-------"+root.val);
}
else
System.out.println(root.val);
printBinaryTree(root.left, level+1);
}
Place this function with your own specified TreeNode and keep the level initially 0, and enjoy!
Here are some of the sample outputs:
| | |-------11
| |-------10
| | |-------9
|-------8
| | |-------7
| |-------6
| | |-------5
4
| |-------3
|-------2
| |-------1
| | | |-------10
| | |-------9
| |-------8
| | |-------7
|-------6
| |-------5
4
| |-------3
|-------2
| |-------1
Only problem is with the extending branches; I will try to solve the problem as soon as possible but till then you can use it too.

Your tree will need twice the distance for each layer:
a
/ \
/ \
/ \
/ \
b c
/ \ / \
/ \ / \
d e f g
/ \ / \ / \ / \
h i j k l m n o
You can save your tree in an array of arrays, one array for every depth:
[[a],[b,c],[d,e,f,g],[h,i,j,k,l,m,n,o]]
If your tree is not full, you need to include empty values in that array:
a
/ \
/ \
/ \
/ \
b c
/ \ / \
/ \ / \
d e f g
/ \ \ / \ \
h i k l m o
[[a],[b,c],[d,e,f,g],[h,i, ,k,l,m, ,o]]
Then you can iterate over the array to print your tree, printing spaces before the first element and between the elements depending on the depth and printing the lines depending on if the corresponding elements in the array for the next layer are filled or not.
If your values can be more than one character long, you need to find the longest value while creating the array representation and multiply all widths and the number of lines accordingly.

Based on VasyaNovikov answer. Improved with some Java magic: Generics and Functional interface.
/**
* Print a tree structure in a pretty ASCII fromat.
* #param prefix Currnet previx. Use "" in initial call!
* #param node The current node. Pass the root node of your tree in initial call.
* #param getChildrenFunc A {#link Function} that returns the children of a given node.
* #param isTail Is node the last of its sibblings. Use true in initial call. (This is needed for pretty printing.)
* #param <T> The type of your nodes. Anything that has a toString can be used.
*/
private <T> void printTreeRec(String prefix, T node, Function<T, List<T>> getChildrenFunc, boolean isTail) {
String nodeName = node.toString();
String nodeConnection = isTail ? "└── " : "├── ";
log.debug(prefix + nodeConnection + nodeName);
List<T> children = getChildrenFunc.apply(node);
for (int i = 0; i < children.size(); i++) {
String newPrefix = prefix + (isTail ? " " : "│ ");
printTreeRec(newPrefix, children.get(i), getChildrenFunc, i == children.size()-1);
}
}
Example initial call:
Function<ChecksumModel, List<ChecksumModel>> getChildrenFunc = node -> getChildrenOf(node)
printTreeRec("", rootNode, getChildrenFunc, true);
Will output something like
└── rootNode
├── childNode1
├── childNode2
│ ├── childNode2.1
│ ├── childNode2.2
│ └── childNode2.3
├── childNode3
└── childNode4

public void printPreety() {
List<TreeNode> list = new ArrayList<TreeNode>();
list.add(head);
printTree(list, getHeight(head));
}
public int getHeight(TreeNode head) {
if (head == null) {
return 0;
} else {
return 1 + Math.max(getHeight(head.left), getHeight(head.right));
}
}
/**
* pass head node in list and height of the tree
*
* #param levelNodes
* #param level
*/
private void printTree(List<TreeNode> levelNodes, int level) {
List<TreeNode> nodes = new ArrayList<TreeNode>();
//indentation for first node in given level
printIndentForLevel(level);
for (TreeNode treeNode : levelNodes) {
//print node data
System.out.print(treeNode == null?" ":treeNode.data);
//spacing between nodes
printSpacingBetweenNodes(level);
//if its not a leaf node
if(level>1){
nodes.add(treeNode == null? null:treeNode.left);
nodes.add(treeNode == null? null:treeNode.right);
}
}
System.out.println();
if(level>1){
printTree(nodes, level-1);
}
}
private void printIndentForLevel(int level){
for (int i = (int) (Math.pow(2,level-1)); i >0; i--) {
System.out.print(" ");
}
}
private void printSpacingBetweenNodes(int level){
//spacing between nodes
for (int i = (int) ((Math.pow(2,level-1))*2)-1; i >0; i--) {
System.out.print(" ");
}
}
Prints Tree in following format:
4
3 7
1 5 8
2 10
9

private StringBuilder prettyPrint(Node root, int currentHeight, int totalHeight) {
StringBuilder sb = new StringBuilder();
int spaces = getSpaceCount(totalHeight-currentHeight + 1);
if(root == null) {
//create a 'spatial' block and return it
String row = String.format("%"+(2*spaces+1)+"s%n", "");
//now repeat this row space+1 times
String block = new String(new char[spaces+1]).replace("\0", row);
return new StringBuilder(block);
}
if(currentHeight==totalHeight) return new StringBuilder(root.data+"");
int slashes = getSlashCount(totalHeight-currentHeight +1);
sb.append(String.format("%"+(spaces+1)+"s%"+spaces+"s", root.data+"", ""));
sb.append("\n");
//now print / and \
// but make sure that left and right exists
char leftSlash = root.left == null? ' ':'/';
char rightSlash = root.right==null? ' ':'\\';
int spaceInBetween = 1;
for(int i=0, space = spaces-1; i<slashes; i++, space --, spaceInBetween+=2) {
for(int j=0; j<space; j++) sb.append(" ");
sb.append(leftSlash);
for(int j=0; j<spaceInBetween; j++) sb.append(" ");
sb.append(rightSlash+"");
for(int j=0; j<space; j++) sb.append(" ");
sb.append("\n");
}
//sb.append("\n");
//now get string representations of left and right subtrees
StringBuilder leftTree = prettyPrint(root.left, currentHeight+1, totalHeight);
StringBuilder rightTree = prettyPrint(root.right, currentHeight+1, totalHeight);
// now line by line print the trees side by side
Scanner leftScanner = new Scanner(leftTree.toString());
Scanner rightScanner = new Scanner(rightTree.toString());
// spaceInBetween+=1;
while(leftScanner.hasNextLine()) {
if(currentHeight==totalHeight-1) {
sb.append(String.format("%-2s %2s", leftScanner.nextLine(), rightScanner.nextLine()));
sb.append("\n");
spaceInBetween-=2;
}
else {
sb.append(leftScanner.nextLine());
sb.append(" ");
sb.append(rightScanner.nextLine()+"\n");
}
}
return sb;
}
private int getSpaceCount(int height) {
return (int) (3*Math.pow(2, height-2)-1);
}
private int getSlashCount(int height) {
if(height <= 3) return height -1;
return (int) (3*Math.pow(2, height-3)-1);
}
https://github.com/murtraja/java-binary-tree-printer
only works for 1 to 2 digit integers (i was lazy to make it generic)

This was the simplest solution for horizontal view. Tried with bunch of examples. Works well for my purpose. Updated from #nitin-k 's answer.
public void print(String prefix, BTNode n, boolean isLeft) {
if (n != null) {
print(prefix + " ", n.right, false);
System.out.println (prefix + ("|-- ") + n.data);
print(prefix + " ", n.left, true);
}
}
Call:
bst.print("", bst.root, false);
Solution:
|-- 80
|-- 70
|-- 60
|-- 50
|-- 40
|-- 30
|-- 20
|-- 10

This is a very simple solution to print out a tree. It is not that pretty, but it is really simple:
enum { kWidth = 6 };
void PrintSpace(int n)
{
for (int i = 0; i < n; ++i)
printf(" ");
}
void PrintTree(struct Node * root, int level)
{
if (!root) return;
PrintTree(root->right, level + 1);
PrintSpace(level * kWidth);
printf("%d", root->data);
PrintTree(root->left, level + 1);
}
Sample output:
106
105
104
103
102
101
100

I needed to print a binary tree in one of my projects, for that I have prepared a java class TreePrinter, one of the sample output is:
[+]
/ \
/ \
/ \
/ \
/ \
[*] \
/ \ [-]
[speed] [2] / \
[45] [12]
Here is the code for class TreePrinter along with class TextNode. For printing any tree you can just create an equivalent tree with TextNode class.
import java.util.ArrayList;
public class TreePrinter {
public TreePrinter(){
}
public static String TreeString(TextNode root){
ArrayList layers = new ArrayList();
ArrayList bottom = new ArrayList();
FillBottom(bottom, root); DrawEdges(root);
int height = GetHeight(root);
for(int i = 0; i s.length()) min = s.length();
if(!n.isEdge) s += "[";
s += n.text;
if(!n.isEdge) s += "]";
layers.set(n.depth, s);
}
StringBuilder sb = new StringBuilder();
for(int i = 0; i temp = new ArrayList();
for(int i = 0; i 0) temp.get(i-1).left = x;
temp.add(x);
}
temp.get(count-1).left = n.left;
n.left.depth = temp.get(count-1).depth+1;
n.left = temp.get(0);
DrawEdges(temp.get(count-1).left);
}
if(n.right != null){
int count = n.right.x - (n.x + n.text.length() + 2);
ArrayList temp = new ArrayList();
for(int i = 0; i 0) temp.get(i-1).right = x;
temp.add(x);
}
temp.get(count-1).right = n.right;
n.right.depth = temp.get(count-1).depth+1;
n.right = temp.get(0);
DrawEdges(temp.get(count-1).right);
}
}
private static void FillBottom(ArrayList bottom, TextNode n){
if(n == null) return;
FillBottom(bottom, n.left);
if(!bottom.isEmpty()){
int i = bottom.size()-1;
while(bottom.get(i).isEdge) i--;
TextNode last = bottom.get(i);
if(!n.isEdge) n.x = last.x + last.text.length() + 3;
}
bottom.add(n);
FillBottom(bottom, n.right);
}
private static boolean isLeaf(TextNode n){
return (n.left == null && n.right == null);
}
private static int GetHeight(TextNode n){
if(n == null) return 0;
int l = GetHeight(n.left);
int r = GetHeight(n.right);
return Math.max(l, r) + 1;
}
}
class TextNode {
public String text;
public TextNode parent, left, right;
public boolean isEdge;
public int x, depth;
public TextNode(String text){
this.text = text;
parent = null; left = null; right = null;
isEdge = false;
x = 0; depth = 0;
}
}
Finally here is a test class for printing given sample:
public class Test {
public static void main(String[] args){
TextNode root = new TextNode("+");
root.left = new TextNode("*"); root.left.parent = root;
root.right = new TextNode("-"); root.right.parent = root;
root.left.left = new TextNode("speed"); root.left.left.parent = root.left;
root.left.right = new TextNode("2"); root.left.right.parent = root.left;
root.right.left = new TextNode("45"); root.right.left.parent = root.right;
root.right.right = new TextNode("12"); root.right.right.parent = root.right;
System.out.println(TreePrinter.TreeString(root));
}
}

https://github.com/AharonSambol/PrettyPrintTreeJava
I know I'm late.. But I made this solution which works not only for simple trees but also for more complex ones (such as multi-lined strings)
Example output:

Print in Console:
500
700 300
200 400
Simple code :
public int getHeight()
{
if(rootNode == null) return -1;
return getHeight(rootNode);
}
private int getHeight(Node node)
{
if(node == null) return -1;
return Math.max(getHeight(node.left), getHeight(node.right)) + 1;
}
public void printBinaryTree(Node rootNode)
{
Queue<Node> rootsQueue = new LinkedList<Node>();
Queue<Node> levelQueue = new LinkedList<Node>();
levelQueue.add(rootNode);
int treeHeight = getHeight();
int firstNodeGap;
int internalNodeGap;
int copyinternalNodeGap;
while(true)
{
System.out.println("");
internalNodeGap = (int)(Math.pow(2, treeHeight + 1) -1);
copyinternalNodeGap = internalNodeGap;
firstNodeGap = internalNodeGap/2;
boolean levelFirstNode = true;
while(!levelQueue.isEmpty())
{
internalNodeGap = copyinternalNodeGap;
Node currNode = levelQueue.poll();
if(currNode != null)
{
if(levelFirstNode)
{
while(firstNodeGap > 0)
{
System.out.format("%s", " ");
firstNodeGap--;
}
levelFirstNode =false;
}
else
{
while(internalNodeGap>0)
{
internalNodeGap--;
System.out.format("%s", " ");
}
}
System.out.format("%3d",currNode.data);
rootsQueue.add(currNode);
}
}
--treeHeight;
while(!rootsQueue.isEmpty())
{
Node currNode = rootsQueue.poll();
if(currNode != null)
{
levelQueue.add(currNode.left);
levelQueue.add(currNode.right);
}
}
if(levelQueue.isEmpty()) break;
}
}

Here's a very versatile tree printer. Not the best looking, but it handles a lot of cases. Feel free to add slashes if you can figure that out.
package com.tomac120.NodePrinter;
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
/**
* Created by elijah on 6/28/16.
*/
public class NodePrinter{
final private List<List<PrintableNodePosition>> nodesByRow;
int maxColumnsLeft = 0;
int maxColumnsRight = 0;
int maxTitleLength = 0;
String sep = " ";
int depth = 0;
public NodePrinter(PrintableNode rootNode, int chars_per_node){
this.setDepth(rootNode,1);
nodesByRow = new ArrayList<>(depth);
this.addNode(rootNode._getPrintableNodeInfo(),0,0);
for (int i = 0;i<chars_per_node;i++){
//sep += " ";
}
}
private void setDepth(PrintableNode info, int depth){
if (depth > this.depth){
this.depth = depth;
}
if (info._getLeftChild() != null){
this.setDepth(info._getLeftChild(),depth+1);
}
if (info._getRightChild() != null){
this.setDepth(info._getRightChild(),depth+1);
}
}
private void addNode(PrintableNodeInfo node, int level, int position){
if (position < 0 && -position > maxColumnsLeft){
maxColumnsLeft = -position;
}
if (position > 0 && position > maxColumnsRight){
maxColumnsRight = position;
}
if (node.getTitleLength() > maxTitleLength){
maxTitleLength = node.getTitleLength();
}
List<PrintableNodePosition> row = this.getRow(level);
row.add(new PrintableNodePosition(node, level, position));
level++;
int depthToUse = Math.min(depth,6);
int levelToUse = Math.min(level,6);
int offset = depthToUse - levelToUse-1;
offset = (int)(Math.pow(offset,Math.log(depthToUse)*1.4));
offset = Math.max(offset,3);
PrintableNodeInfo leftChild = node.getLeftChildInfo();
PrintableNodeInfo rightChild = node.getRightChildInfo();
if (leftChild != null){
this.addNode(leftChild,level,position-offset);
}
if (rightChild != null){
this.addNode(rightChild,level,position+offset);
}
}
private List<PrintableNodePosition> getRow(int row){
if (row > nodesByRow.size() - 1){
nodesByRow.add(new LinkedList<>());
}
return nodesByRow.get(row);
}
public void print(){
int max_chars = this.maxColumnsLeft+maxColumnsRight+1;
int level = 0;
String node_format = "%-"+this.maxTitleLength+"s";
for (List<PrintableNodePosition> pos_arr : this.nodesByRow){
String[] chars = this.getCharactersArray(pos_arr,max_chars);
String line = "";
int empty_chars = 0;
for (int i=0;i<chars.length+1;i++){
String value_i = i < chars.length ? chars[i]:null;
if (chars.length + 1 == i || value_i != null){
if (empty_chars > 0) {
System.out.print(String.format("%-" + empty_chars + "s", " "));
}
if (value_i != null){
System.out.print(String.format(node_format,value_i));
empty_chars = -1;
} else{
empty_chars = 0;
}
} else {
empty_chars++;
}
}
System.out.print("\n");
int depthToUse = Math.min(6,depth);
int line_offset = depthToUse - level;
line_offset *= 0.5;
line_offset = Math.max(0,line_offset);
for (int i=0;i<line_offset;i++){
System.out.println("");
}
level++;
}
}
private String[] getCharactersArray(List<PrintableNodePosition> nodes, int max_chars){
String[] positions = new String[max_chars+1];
for (PrintableNodePosition a : nodes){
int pos_i = maxColumnsLeft + a.column;
String title_i = a.nodeInfo.getTitleFormatted(this.maxTitleLength);
positions[pos_i] = title_i;
}
return positions;
}
}
NodeInfo class
package com.tomac120.NodePrinter;
/**
* Created by elijah on 6/28/16.
*/
public class PrintableNodeInfo {
public enum CLI_PRINT_COLOR {
RESET("\u001B[0m"),
BLACK("\u001B[30m"),
RED("\u001B[31m"),
GREEN("\u001B[32m"),
YELLOW("\u001B[33m"),
BLUE("\u001B[34m"),
PURPLE("\u001B[35m"),
CYAN("\u001B[36m"),
WHITE("\u001B[37m");
final String value;
CLI_PRINT_COLOR(String value){
this.value = value;
}
#Override
public String toString() {
return value;
}
}
private final String title;
private final PrintableNode leftChild;
private final PrintableNode rightChild;
private final CLI_PRINT_COLOR textColor;
public PrintableNodeInfo(String title, PrintableNode leftChild, PrintableNode rightChild){
this(title,leftChild,rightChild,CLI_PRINT_COLOR.BLACK);
}
public PrintableNodeInfo(String title, PrintableNode leftChild, PrintableNode righthild, CLI_PRINT_COLOR textColor){
this.title = title;
this.leftChild = leftChild;
this.rightChild = righthild;
this.textColor = textColor;
}
public String getTitle(){
return title;
}
public CLI_PRINT_COLOR getTextColor(){
return textColor;
}
public String getTitleFormatted(int max_chars){
return this.textColor+title+CLI_PRINT_COLOR.RESET;
/*
String title = this.title.length() > max_chars ? this.title.substring(0,max_chars+1):this.title;
boolean left = true;
while(title.length() < max_chars){
if (left){
title = " "+title;
} else {
title = title + " ";
}
}
return this.textColor+title+CLI_PRINT_COLOR.RESET;*/
}
public int getTitleLength(){
return title.length();
}
public PrintableNodeInfo getLeftChildInfo(){
if (leftChild == null){
return null;
}
return leftChild._getPrintableNodeInfo();
}
public PrintableNodeInfo getRightChildInfo(){
if (rightChild == null){
return null;
}
return rightChild._getPrintableNodeInfo();
}
}
NodePosition class
package com.tomac120.NodePrinter;
/**
* Created by elijah on 6/28/16.
*/
public class PrintableNodePosition implements Comparable<PrintableNodePosition> {
public final int row;
public final int column;
public final PrintableNodeInfo nodeInfo;
public PrintableNodePosition(PrintableNodeInfo nodeInfo, int row, int column){
this.row = row;
this.column = column;
this.nodeInfo = nodeInfo;
}
#Override
public int compareTo(PrintableNodePosition o) {
return Integer.compare(this.column,o.column);
}
}
And, finally, Node Interface
package com.tomac120.NodePrinter;
/**
* Created by elijah on 6/28/16.
*/
public interface PrintableNode {
PrintableNodeInfo _getPrintableNodeInfo();
PrintableNode _getLeftChild();
PrintableNode _getRightChild();
}

A Scala solution, adapted from Vasya Novikov's answer and specialized for binary trees:
/** An immutable Binary Tree. */
case class BTree[T](value: T, left: Option[BTree[T]], right: Option[BTree[T]]) {
/* Adapted from: http://stackoverflow.com/a/8948691/643684 */
def pretty: String = {
def work(tree: BTree[T], prefix: String, isTail: Boolean): String = {
val (line, bar) = if (isTail) ("└── ", " ") else ("├── ", "│")
val curr = s"${prefix}${line}${tree.value}"
val rights = tree.right match {
case None => s"${prefix}${bar} ├── ∅"
case Some(r) => work(r, s"${prefix}${bar} ", false)
}
val lefts = tree.left match {
case None => s"${prefix}${bar} └── ∅"
case Some(l) => work(l, s"${prefix}${bar} ", true)
}
s"${curr}\n${rights}\n${lefts}"
}
work(this, "", true)
}
}

Here is another way to visualize your tree: save the nodes as an xml file and then let your browser show you the hierarchy:
class treeNode{
int key;
treeNode left;
treeNode right;
public treeNode(int key){
this.key = key;
left = right = null;
}
public void printNode(StringBuilder output, String dir){
output.append("<node key='" + key + "' dir='" + dir + "'>");
if(left != null)
left.printNode(output, "l");
if(right != null)
right.printNode(output, "r");
output.append("</node>");
}
}
class tree{
private treeNode treeRoot;
public tree(int key){
treeRoot = new treeNode(key);
}
public void insert(int key){
insert(treeRoot, key);
}
private treeNode insert(treeNode root, int key){
if(root == null){
treeNode child = new treeNode(key);
return child;
}
if(key < root.key)
root.left = insert(root.left, key);
else if(key > root.key)
root.right = insert(root.right, key);
return root;
}
public void saveTreeAsXml(){
StringBuilder strOutput = new StringBuilder();
strOutput.append("<?xml version=\"1.0\" encoding=\"UTF-8\"?>");
treeRoot.printNode(strOutput, "root");
try {
PrintWriter writer = new PrintWriter("C:/tree.xml", "UTF-8");
writer.write(strOutput.toString());
writer.close();
}
catch (FileNotFoundException e){
}
catch(UnsupportedEncodingException e){
}
}
}
Here is code to test it:
tree t = new tree(1);
t.insert(10);
t.insert(5);
t.insert(4);
t.insert(20);
t.insert(40);
t.insert(30);
t.insert(80);
t.insert(60);
t.insert(50);
t.saveTreeAsXml();
And the output looks like this:

using map...
{
Map<Integer,String> m = new LinkedHashMap<>();
tn.printNodeWithLvl(node,l,m);
for(Entry<Integer, String> map :m.entrySet()) {
System.out.println(map.getValue());
}
then....method
private void printNodeWithLvl(Node node,int l,Map<Integer,String> m) {
if(node==null) {
return;
}
if(m.containsKey(l)) {
m.put(l, new StringBuilder(m.get(l)).append(node.value).toString());
}else {
m.put(l, node.value+"");
}
l++;
printNodeWithLvl( node.left,l,m);
printNodeWithLvl(node.right,l,m);
}
}

Horizontal representation is a little complex compared to vertical representation. Vertical printing is just plain RNL(Right->Node->left or mirror of inorder) traversal so that right subtree is printed first then left subtree.
def printFullTree(root, delim=' ', idnt=[], left=None):
if root:
idnt.append(delim)
x, y = setDelims(left)
printFullTree(root.right, x, idnt, False)
indent2(root.val, idnt)
printFullTree(root.left, y, idnt, True)
idnt.pop()
def setDelims(left):
x = ' '; y='|'
return (y,x) if (left == True) else (x,y) if (left == False) else (x,x)
def indent2(x, idnt, width=6):
for delim in idnt:
print(delim + ' '*(width-1), end='')
print('|->', x)
output:
|-> 15
|-> 14
| |-> 13
|-> 12
| | |-> 11
| |-> 10
| |-> 9
|-> 8
| |-> 7
| |-> 6
| | |-> 4
|-> 3
| |-> 2
|-> 1
|-> 0
In horizontal representation, the display is built by HashMap of TreeMap or HashMap<Integer, TreeMap<Integer, Object>> xy; where HashMap contains node's y-axis/level_no as Key and TreeMap as value. The Treemap interally holds all nodes in same level, sorted by their x axis value as key starting from leftmost -ve, root=0, rightmost=+ve.
Using HashMap makes algo work in O(1) lookup for each level and TreeMap for sorted order in O(logn) if self balancing tree/Treap is used.
Still while doing so don't forget to store placeholders for null child such as ' '/spaces so that the tree looks as intended.
Now the only thing left is to calculate horizontal node distance, this can be done with some math calc,
calc tree width and height.
once done, when displaying the nodes, present them at a optimal distance based on calculated width, height, and skew info if any.

Try this:
public static void print(int[] minHeap, int minWidth) {
int size = minHeap.length;
int level = log2(size);
int maxLength = (int) Math.pow(2, level) * minWidth;
int currentLevel = -1 ;
int width = maxLength;
for (int i = 0; i < size; i++) {
if (log2(i + 1) > currentLevel) {
currentLevel++;
System.out.println();
width = maxLength / (int) Math.pow(2, currentLevel);
}
System.out.print(StringUtils.center(String.valueOf(minHeap[i]), width));
}
System.out.println();
}
private static int log2(int n) {
return (int) (Math.log(n) / Math.log(2));
}
The idea of this code snippet is dividing the maxLength which is the length of bottom line by the number of elements of each line to get the block width. Then put the element in the middle of each block.
The parameter minWidth implys the length of blocks in the bottom line.
A picture to illustrate the idea and show the result.