class Solution {
public List<List<Integer>> pathSum(TreeNode root, int sum) {
List<List<Integer>> res = new ArrayList<>();
if(root == null) return res;
List<Integer> list = new ArrayList<>();
helper(res, list, root, sum);
return res;
}
public void helper(List<List<Integer>> res, List<Integer> list, TreeNode root, int sum){
list.add(root.val);
if(root.left == null && root.right == null){
if(root.val == sum)
res.add(new ArrayList<>(list));
}
if(root.left != null)
helper(res, list, root.left, sum-root.val);
if(root.right != null)
helper(res, list, root.right, sum-root.val);
list.remove(list.size()-1);
}
}
I am working on Leetcode 113. Path Sum II. It's not about the problem itself. I want to know why I have to write res.add(new ArrayList<>(list)); instead of writing res.add(list); directly in line 13.
Writing new ArrayList<>(list) creates a new list, containing all the elements in list. This is required, since at the end of the function you are calling list.remove(list.size()-1);, you are thus modifying the list variable.
If you were to add list directly in res, the remove call would also modify res.
Another relevant example:
class MyClass {
public int modify = 5;
}
class Test {
public static void myFunction() {
MyClass object = new MyClass();
System.out.println(object.modify); // prints 5.
ArrayList<MyClass> myList = new ArrayList<>();
myList.add(object);
object.modify = 800;
for(MyClass item : myList) {
System.out.println(item.modify); // prints 800.
}
}
}
new ArrayList<>(list) will make a copy of list, which will make sure the elements be stored into res.
Otherwise, after list.remove(list.size()-1), the elements in res will be removed.
I was doing this exercice:
Write code to partition a linked list around a value x, such that all nodes less than x come before all nodes greater than or equal to x. Example input: 3 -> 5 -> 8 -> 5 -> 10 -> 2 -> 1 output: 3 -> 1 -> 2 -> 10 -> 5 -> 5 -> 8
I found it hard to find a solution for Singly linked list (that created by my own, not using library), I would like to know if there is uncessary code blocks in my code and is there a way to avoid putting in two lists and then merge? because it seems to have very slow performance like that.
public CustomLinkedList partition(CustomLinkedList list, int x) {
CustomLinkedList beforeL = new CustomLinkedList();
CustomLinkedList afterL = new CustomLinkedList();
LinkedListNode current = list.getHead();
while (current != null) {
if (current.getData() < x) {
addToLinkedList(beforeL, current.getData());
} else {
addToLinkedList(afterL, current.getData());
}
// increment current
current = current.getNext();
}
if (beforeL.getHead() == null)
return afterL;
mergeLinkedLists(beforeL, afterL);
return beforeL;
}
public void addToLinkedList(CustomLinkedList list, int value) {
LinkedListNode newEnd = new LinkedListNode(value);
LinkedListNode cur = list.getHead();
if (cur == null)
list.setHead(newEnd);
else {
while (cur.getNext() != null) {
cur = cur.getNext();
}
cur.setNext(newEnd);
cur = newEnd;
}
}
public void mergeLinkedLists(CustomLinkedList list1, CustomLinkedList list2) {
LinkedListNode start = list1.getHead();
LinkedListNode prev = null;
while (start != null) {
prev = start;
start = start.getNext();
}
prev.setNext(list2.getHead());
}
CustumLinkedList contains two attributes: -LinkedListNode which is the head and an int which is the size.
LinkedListNode contains two attributes: One of type LinkedListNode pointing to next node and one of type int: data value
Thank you.
The problem of your code is not merging two lists as you mentioned. It's wrong to use the word merge here because you're only linking up the tail of the left list with head of right list which is a constant time operation.
The real problem is - on inserting a new element on the left or right list, you are iterating from head to tail every time which yields in-total O(n^2) operation and is definitely slow.
Here I've wrote a simpler version and avoid iterating every time from head to insert a new item by keeping track of the current tail.
The code is very simple and is definitely faster than yours(O(n)). Let me know if you need explanation on any part.
// I don't know how your CustomLinkedList is implemented. Here I wrote a simple LinkedList node
public class ListNode {
private int val;
private ListNode next;
public ListNode(int x) {
val = x;
}
public int getValue() {
return this.val;
}
public ListNode getNext() {
return this.next;
}
public void setNext(ListNode next) {
this.next = next;
}
}
public ListNode partition(ListNode head, int x) {
if(head == null) return null;
ListNode left = null;
ListNode right = null;
ListNode iterL = left;
ListNode iterR = right;
while(iter != null) {
if(iter.getValue() < x) {
iterL = addNode(iterL, iter.getValue());
}
else {
iterR = addNode(iterR, iter.getValue());
}
iter = iter.getNext();
}
// link up the left and right list
iterL.setNext(iterR);
return left;
}
public ListNode addNode(ListNode curr, int value) {
ListNode* newNode = new ListNode(value);
if(curr == null) {
curr = newNode;
} else {
curr.setNext(newNode);
curr = curr.getNext();
}
return curr;
}
Hope it helps!
If you have any list of data, access orderByX Method.
Hope it would help you.
public class OrderByX {
Nodes root = null;
OrderByX() {
root = null;
}
void create(int[] array, int k) {
for (int i = 0; i < array.length; ++i) {
root = insert(root, array[i]);
}
}
Nodes insert(Nodes root, int data) {
if (root == null) {
root = new Nodes(data);
} else {
Nodes tempNew = new Nodes(data);
tempNew.setNext(root);
root = tempNew;
}
return root;
}
void display() {
Nodes tempNode = root;
while (tempNode != null) {
System.out.print(tempNode.getData() + ", ");
tempNode = tempNode.getNext();
}
}
void displayOrder(Nodes root) {
if (root == null) {
return;
} else {
displayOrder(root.getNext());
System.out.print(root.getData() + ", ");
}
}
Nodes orderByX(Nodes root, int x) {
Nodes resultNode = null;
Nodes lessNode = null;
Nodes greatNode = null;
Nodes midNode = null;
while (root != null) {
if (root.getData() < x) {
if (lessNode == null) {
lessNode = root;
root = root.getNext();
lessNode.setNext(null);
} else {
Nodes temp = root.getNext();
root.setNext(lessNode);
lessNode = root;
root = temp;
}
} else if (root.getData() > x) {
if (greatNode == null) {
greatNode = root;
root = root.getNext();
greatNode.setNext(null);
} else {
Nodes temp = root.getNext();
root.setNext(greatNode);
greatNode = root;
root = temp;
}
} else {
if (midNode == null) {
midNode = root;
root = root.getNext();
midNode.setNext(null);
} else {
Nodes temp = root.getNext();
root.setNext(midNode);
midNode = root;
root = temp;
}
}
}
resultNode = lessNode;
while (lessNode.getNext() != null) {
lessNode = lessNode.getNext();
}
lessNode.setNext(midNode);
while (midNode.getNext() != null) {
midNode = midNode.getNext();
}
midNode.setNext(greatNode);
return resultNode;
}
public static void main(String... args) {
int[] array = { 7, 1, 6, 2, 8 };
OrderByX obj = new OrderByX();
obj.create(array, 0);
obj.display();
System.out.println();
obj.displayOrder(obj.root);
System.out.println();
obj.root = obj.orderByX(obj.root, 2);
obj.display();
}
}
class Nodes {
private int data;
private Nodes next;
Nodes(int data) {
this.data = data;
this.next = null;
}
public Nodes getNext() {
return next;
}
public void setNext(Nodes next) {
this.next = next;
}
public int getData() {
return data;
}
public void setData(int data) {
this.data = data;
}
}
I think that maintaining two lists is not an issue. It is possible to use a single list, but at the cost of loosing some of the simplicity.
The principal problem seems to be the addToLinkedList(CustomLinkedList list, int value) method.
It iterates throughout the entire list in order to add a new element.
One alternative is to always add elements at the front of the list. This would also produce a valid solution, and would run faster.
Okay, I have read through all the other related questions and cannot find one that helps with java. I get the general idea from deciphering what i can in other languages; but i am yet to figure it out.
Problem: I would like to level sort (which i have working using recursion) and print it out in the general shape of a tree.
So say i have this:
1
/ \
2 3
/ / \
4 5 6
My code prints out the level order like this:
1 2 3 4 5 6
I want to print it out like this:
1
2 3
4 5 6
Now before you give me a moral speech about doing my work... I have already finished my AP Comp Sci project and got curious about this when my teacher mentioned the Breadth First Search thing.
I don't know if it will help, but here is my code so far:
/**
* Calls the levelOrder helper method and prints out in levelOrder.
*/
public void levelOrder()
{
q = new QueueList();
treeHeight = height();
levelOrder(myRoot, q, myLevel);
}
/**
* Helper method that uses recursion to print out the tree in
* levelOrder
*/
private void levelOrder(TreeNode root, QueueList q, int curLev)
{
System.out.print(curLev);
if(root == null)
{
return;
}
if(q.isEmpty())
{
System.out.println(root.getValue());
}
else
{
System.out.print((String)q.dequeue()+", ");
}
if(root.getLeft() != null)
{
q.enqueue(root.getLeft().getValue());
System.out.println();
}
if(root.getRight() != null)
{
q.enqueue(root.getRight().getValue());
System.out.println();
curLev++;
}
levelOrder(root.getLeft(),q, curLev);
levelOrder(root.getRight(),q, curLev);
}
From what i can figure out, i will need to use the total height of the tree, and use a level counter... Only problem is my level counter keeps counting when my levelOrder uses recursion to go back through the tree.
Sorry if this is to much, but some tips would be nice. :)
Here is the code, this question was asked to me in one of the interviews...
public void printTree(TreeNode tmpRoot) {
Queue<TreeNode> currentLevel = new LinkedList<TreeNode>();
Queue<TreeNode> nextLevel = new LinkedList<TreeNode>();
currentLevel.add(tmpRoot);
while (!currentLevel.isEmpty()) {
Iterator<TreeNode> iter = currentLevel.iterator();
while (iter.hasNext()) {
TreeNode currentNode = iter.next();
if (currentNode.left != null) {
nextLevel.add(currentNode.left);
}
if (currentNode.right != null) {
nextLevel.add(currentNode.right);
}
System.out.print(currentNode.value + " ");
}
System.out.println();
currentLevel = nextLevel;
nextLevel = new LinkedList<TreeNode>();
}
}
This is the easiest solution
public void byLevel(Node root){
Queue<Node> level = new LinkedList<>();
level.add(root);
while(!level.isEmpty()){
Node node = level.poll();
System.out.print(node.item + " ");
if(node.leftChild!= null)
level.add(node.leftChild);
if(node.rightChild!= null)
level.add(node.rightChild);
}
}
https://github.com/camluca/Samples/blob/master/Tree.java
in my github you can find other helpful functions in the class Tree like:
Displaying the tree
****......................................................****
42
25 65
12 37 43 87
9 13 30 -- -- -- -- 99
****......................................................****
Inorder traversal
9 12 13 25 30 37 42 43 65 87 99
Preorder traversal
42 25 12 9 13 37 30 65 43 87 99
Postorder traversal
9 13 12 30 37 25 43 99 87 65 42
By Level
42 25 65 12 37 43 87 9 13 30 99
Here is how I would do it:
levelOrder(List<TreeNode> n) {
List<TreeNode> next = new List<TreeNode>();
foreach(TreeNode t : n) {
print(t);
next.Add(t.left);
next.Add(t.right);
}
println();
levelOrder(next);
}
(Was originally going to be real code - got bored partway through, so it's psueodocodey)
Just thought of sharing Anon's suggestion in real java code and fixing a couple of KEY issues (like there is not an end condition for the recursion so it never stops adding to the stack, and not checking for null in the received array gets you a null pointer exception).
Also there is no exception as Eric Hauser suggests, because it is not modifying the collection its looping through, it's modifying a new one.
Here it goes:
public void levelOrder(List<TreeNode> n) {
List<TreeNode> next = new ArrayList<TreeNode>();
for (TreeNode t : n) {
if (t != null) {
System.out.print(t.getValue());
next.add(t.getLeftChild());
next.add(t.getRightChild());
}
}
System.out.println();
if(next.size() > 0)levelOrder(next);
}
Below method returns ArrayList of ArrayList containing all nodes level by level:-
public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) {
ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
if(root == null) return result;
Queue q1 = new LinkedList();
Queue q2 = new LinkedList();
ArrayList<Integer> list = new ArrayList<Integer>();
q1.add(root);
while(!q1.isEmpty() || !q2.isEmpty()){
while(!q1.isEmpty()){
TreeNode temp = (TreeNode)q1.poll();
list.add(temp.val);
if(temp.left != null) q2.add(temp.left);
if(temp.right != null) q2.add(temp.right);
}
if(list.size() > 0)result.add(new ArrayList<Integer>(list));
list.clear();
while(!q2.isEmpty()){
TreeNode temp = (TreeNode)q2.poll();
list.add(temp.val);
if(temp.left != null) q1.add(temp.left);
if(temp.right != null) q1.add(temp.right);
}
if(list.size() > 0)result.add(new ArrayList<Integer>(list));
list.clear();
}
return result;
}
The answer is close....the only issue I could see with it is that if a tree doesn't have a node in a particular position, you would set that pointer to null. What happens when you try to put a null pointer into the list?
Here is something I did for a recent assignment. It works flawlessly. You can use it starting from any root.
//Prints the tree in level order
public void printTree(){
printTree(root);
}
public void printTree(TreeNode tmpRoot){
//If the first node isn't null....continue on
if(tmpRoot != null){
Queue<TreeNode> currentLevel = new LinkedList<TreeNode>(); //Queue that holds the nodes on the current level
Queue<TreeNode> nextLevel = new LinkedList<TreeNode>(); //Queue the stores the nodes for the next level
int treeHeight = height(tmpRoot); //Stores the height of the current tree
int levelTotal = 0; //keeps track of the total levels printed so we don't pass the height and print a billion "null"s
//put the root on the currnt level's queue
currentLevel.add(tmpRoot);
//while there is still another level to print and we haven't gone past the tree's height
while(!currentLevel.isEmpty()&& (levelTotal< treeHeight)){
//Print the next node on the level, add its childen to the next level's queue, and dequeue the node...do this until the current level has been printed
while(!currentLevel.isEmpty()){
//Print the current value
System.out.print(currentLevel.peek().getValue()+" ");
//If there is a left pointer, put the node on the nextLevel's stack. If there is no ponter, add a node with a null value to the next level's stack
tmpRoot = currentLevel.peek().getLeft();
if(tmpRoot != null)
nextLevel.add(tmpRoot);
else
nextLevel.add(new TreeNode(null));
//If there is a right pointer, put the node on the nextLevel's stack. If there is no ponter, add a node with a null value to the next level's stack
tmpRoot = currentLevel.remove().getRight();
if(tmpRoot != null)
nextLevel.add(tmpRoot);
else
nextLevel.add(new TreeNode(null));
}//end while(!currentLevel.isEmpty())
//populate the currentLevel queue with items from the next level
while(!nextLevel.isEmpty()){
currentLevel.add(nextLevel.remove());
}
//Print a blank line to show height
System.out.println("");
//flag that we are working on the next level
levelTotal++;
}//end while(!currentLevel.isEmpty())
}//end if(tmpRoot != null)
}//end method printTree
public int height(){
return height(getRoot());
}
public int height(TreeNode tmpRoot){
if (tmpRoot == null)
return 0;
int leftHeight = height(tmpRoot.getLeft());
int rightHeight = height(tmpRoot.getRight());
if(leftHeight >= rightHeight)
return leftHeight + 1;
else
return rightHeight + 1;
}
I really like the simplicity of Anon's code; its elegant. But, sometimes elegant code doesn't always translate into code that is intuitively easy to grasp. So, here's my attempt to show a similar approach that requires Log(n) more space, but should read more naturally to those who are most familiar with depth first search (going down the length of a tree)
The following snippet of code sets nodes belonging to a particular level in a list, and arranges that list in a list that holds all the levels of the tree. Hence the List<List<BinaryNode<T>>> that you will see below. The rest should be fairly self explanatory.
public static final <T extends Comparable<T>> void printTreeInLevelOrder(
BinaryTree<T> tree) {
BinaryNode<T> root = tree.getRoot();
List<List<BinaryNode<T>>> levels = new ArrayList<List<BinaryNode<T>>>();
addNodesToLevels(root, levels, 0);
for(List<BinaryNode<T>> level: levels){
for(BinaryNode<T> node: level){
System.out.print(node+ " ");
}
System.out.println();
}
}
private static final <T extends Comparable<T>> void addNodesToLevels(
BinaryNode<T> node, List<List<BinaryNode<T>>> levels, int level) {
if(null == node){
return;
}
List<BinaryNode<T>> levelNodes;
if(levels.size() == level){
levelNodes = new ArrayList<BinaryNode<T>>();
levels.add(level, levelNodes);
}
else{
levelNodes = levels.get(level);
}
levelNodes.add(node);
addNodesToLevels(node.getLeftChild(), levels, level+1);
addNodesToLevels(node.getRightChild(), levels, level+1);
}
Following implementation uses 2 queues. Using ListBlokcingQueue here but any queue would work.
import java.util.concurrent.*;
public class Test5 {
public class Tree {
private String value;
private Tree left;
private Tree right;
public Tree(String value) {
this.value = value;
}
public void setLeft(Tree t) {
this.left = t;
}
public void setRight(Tree t) {
this.right = t;
}
public Tree getLeft() {
return this.left;
}
public Tree getRight() {
return this.right;
}
public String getValue() {
return this.value;
}
}
Tree tree = null;
public void setTree(Tree t) {
this.tree = t;
}
public void printTree() {
LinkedBlockingQueue<Tree> q = new LinkedBlockingQueue<Tree>();
q.add(this.tree);
while (true) {
LinkedBlockingQueue<Tree> subQueue = new LinkedBlockingQueue<Tree>();
while (!q.isEmpty()) {
Tree aTree = q.remove();
System.out.print(aTree.getValue() + ", ");
if (aTree.getLeft() != null) {
subQueue.add(aTree.getLeft());
}
if (aTree.getRight() != null) {
subQueue.add(aTree.getRight());
}
}
System.out.println("");
if (subQueue.isEmpty()) {
return;
} else {
q = subQueue;
}
}
}
public void testPrint() {
Tree a = new Tree("A");
a.setLeft(new Tree("B"));
a.setRight(new Tree("C"));
a.getLeft().setLeft(new Tree("D"));
a.getLeft().setRight(new Tree("E"));
a.getRight().setLeft(new Tree("F"));
a.getRight().setRight(new Tree("G"));
setTree(a);
printTree();
}
public static void main(String args[]) {
Test5 test5 = new Test5();
test5.testPrint();
}
}
public class PrintATreeLevelByLevel {
public static class Node{
int data;
public Node left;
public Node right;
public Node(int data){
this.data = data;
this.left = null;
this.right = null;
}
}
public void printATreeLevelByLevel(Node n){
Queue<Node> queue = new LinkedList<Node>();
queue.add(n);
int node = 1; //because at root
int child = 0; //initialize it with 0
while(queue.size() != 0){
Node n1 = queue.remove();
node--;
System.err.print(n1.data +" ");
if(n1.left !=null){
queue.add(n1.left);
child ++;
}
if(n1.right != null){
queue.add(n1.right);
child ++;
}
if( node == 0){
System.err.println();
node = child ;
child = 0;
}
}
}
public static void main(String[]args){
PrintATreeLevelByLevel obj = new PrintATreeLevelByLevel();
Node node1 = new Node(1);
Node node2 = new Node(2);
Node node3 = new Node(3);
Node node4 = new Node(4);
Node node5 = new Node(5);
Node node6 = new Node(6);
Node node7 = new Node(7);
Node node8 = new Node(8);
node4.left = node2;
node4.right = node6;
node2.left = node1;
// node2.right = node3;
node6.left = node5;
node6.right = node7;
node1.left = node8;
obj.printATreeLevelByLevel(node4);
}
}
Try this, using 2 Queues to keep track of the levels.
public static void printByLevel(Node root){
LinkedList<Node> curLevel = new LinkedList<Node>();
LinkedList<Node> nextLevel = curLevel;
StringBuilder sb = new StringBuilder();
curLevel.add(root);
sb.append(root.data + "\n");
while(nextLevel.size() > 0){
nextLevel = new LinkedList<Node>();
for (int i = 0; i < curLevel.size(); i++){
Node cur = curLevel.get(i);
if (cur.left != null) {
nextLevel.add(cur.left);
sb.append(cur.left.data + " ");
}
if (cur.right != null) {
nextLevel.add(cur.right);
sb.append(cur.right.data + " ");
}
}
if (nextLevel.size() > 0) {
sb.append("\n");
curLevel = nextLevel;
}
}
System.out.println(sb.toString());
}
A - Solution
I've written direct solution here. If you want the detailed answer, demo code and explanation, you can skip and check the rest headings of the answer;
public static <T> void printLevelOrder(TreeNode<T> root) {
System.out.println("Tree;");
System.out.println("*****");
// null check
if(root == null) {
System.out.printf(" Empty\n");
return;
}
MyQueue<TreeNode<T>> queue = new MyQueue<>();
queue.enqueue(root);
while(!queue.isEmpty()) {
handleLevel(queue);
}
}
// process each level
private static <T> void handleLevel(MyQueue<TreeNode<T>> queue) {
int size = queue.size();
for(int i = 0; i < size; i++) {
TreeNode<T> temp = queue.dequeue();
System.out.printf("%s ", temp.data);
queue.enqueue(temp.left);
queue.enqueue(temp.right);
}
System.out.printf("\n");
}
B - Explanation
In order to print a tree in level-order, you should process each level using a simple queue implementation. In my demo, I've written a very minimalist simple queue class called as MyQueue.
Public method printLevelOrder will take the TreeNode<T> object instance root as a parameter which stands for the root of the tree. The private method handleLevel takes the MyQueue instance as a parameter.
On each level, handleLevel method dequeues the queue as much as the size of the queue. The level restriction is controlled as this process is executed only with the size of the queue which exactly equals to the elements of that level then puts a new line character to the output.
C - TreeNode class
public class TreeNode<T> {
T data;
TreeNode<T> left;
TreeNode<T> right;
public TreeNode(T data) {
this.data = data;;
}
}
D - MyQueue class : A simple Queue Implementation
public class MyQueue<T> {
private static class Node<T> {
T data;
Node next;
public Node(T data) {
this(data, null);
}
public Node(T data, Node<T> next) {
this.data = data;
this.next = next;
}
}
private Node head;
private Node tail;
private int size;
public MyQueue() {
head = null;
tail = null;
}
public int size() {
return size;
}
public void enqueue(T data) {
if(data == null)
return;
if(head == null)
head = tail = new Node(data);
else {
tail.next = new Node(data);
tail = tail.next;
}
size++;
}
public T dequeue() {
if(tail != null) {
T temp = (T) head.data;
head = head.next;
size--;
return temp;
}
return null;
}
public boolean isEmpty() {
return size == 0;
}
public void printQueue() {
System.out.println("Queue: ");
if(head == null)
return;
else {
Node<T> temp = head;
while(temp != null) {
System.out.printf("%s ", temp.data);
temp = temp.next;
}
}
System.out.printf("%n");
}
}
E - DEMO : Printing Tree in Level-Order
public class LevelOrderPrintDemo {
public static void main(String[] args) {
// root level
TreeNode<Integer> root = new TreeNode<>(1);
// level 1
root.left = new TreeNode<>(2);
root.right = new TreeNode<>(3);
// level 2
root.left.left = new TreeNode<>(4);
root.right.left = new TreeNode<>(5);
root.right.right = new TreeNode<>(6);
/*
* 1 root
* / \
* 2 3 level-1
* / / \
* 4 5 6 level-2
*/
printLevelOrder(root);
}
public static <T> void printLevelOrder(TreeNode<T> root) {
System.out.println("Tree;");
System.out.println("*****");
// null check
if(root == null) {
System.out.printf(" Empty\n");
return;
}
MyQueue<TreeNode<T>> queue = new MyQueue<>();
queue.enqueue(root);
while(!queue.isEmpty()) {
handleLevel(queue);
}
}
// process each level
private static <T> void handleLevel(MyQueue<TreeNode<T>> queue) {
int size = queue.size();
for(int i = 0; i < size; i++) {
TreeNode<T> temp = queue.dequeue();
System.out.printf("%s ", temp.data);
queue.enqueue(temp.left);
queue.enqueue(temp.right);
}
System.out.printf("\n");
}
}
F - Sample Input
1 // root
/ \
2 3 // level-1
/ / \
4 5 6 // level-2
G - Sample Output
Tree;
*****
1
2 3
4 5 6
public void printAllLevels(BNode node, int h){
int i;
for(i=1;i<=h;i++){
printLevel(node,i);
System.out.println();
}
}
public void printLevel(BNode node, int level){
if (node==null)
return;
if (level==1)
System.out.print(node.value + " ");
else if (level>1){
printLevel(node.left, level-1);
printLevel(node.right, level-1);
}
}
public int height(BNode node) {
if (node == null) {
return 0;
} else {
return 1 + Math.max(height(node.left),
height(node.right));
}
}
First of all, I do not like to take credit for this solution. It's a modification of somebody's function and I tailored it to provide the solution.
I am using 3 functions here.
First I calculate the height of the tree.
I then have a function to print a particular level of the tree.
Using the height of the tree and the function to print the level of a tree, I traverse the tree and iterate and print all levels of the tree using my third function.
I hope this helps.
EDIT: The time complexity on this solution for printing all node in level order traversal will not be O(n). The reason being, each time you go down a level, you will visit the same nodes again and again.
If you are looking for a O(n) solution, i think using Queues would be a better option.
I think we can achieve this by using one queue itself. This is a java implementation using one queue only. Based on BFS...
public void BFSPrint()
{
Queue<Node> q = new LinkedList<Node>();
q.offer(root);
BFSPrint(q);
}
private void BFSPrint(Queue<Node> q)
{
if(q.isEmpty())
return;
int qLen = q.size(),i=0;
/*limiting it to q size when it is passed,
this will make it print in next lines. if we use iterator instead,
we will again have same output as question, because iterator
will end only q empties*/
while(i<qLen)
{
Node current = q.remove();
System.out.print(current.data+" ");
if(current.left!=null)
q.offer(current.left);
if(current.right!=null)
q.offer(current.right);
i++;
}
System.out.println();
BFSPrint(q);
}
the top solutions only print the children of each node together. This is wrong according to the description.
What we need is all the nodes of the same level together in the same line.
1) Apply BFS
2) Store heights of nodes to a map that will hold level - list of nodes.
3) Iterate over the map and print out the results.
See Java code below:
public void printByLevel(Node root){
Queue<Node> q = new LinkedBlockingQueue<Node>();
root.visited = true;
root.height=1;
q.add(root);
//Node height - list of nodes with same level
Map<Integer, List<Node>> buckets = new HashMap<Integer, List<Node>>();
addToBuckets(buckets, root);
while (!q.isEmpty()){
Node r = q.poll();
if (r.adjacent!=null)
for (Node n : r.adjacent){
if (!n.visited){
n.height = r.height+1; //adjust new height
addToBuckets(buckets, n);
n.visited = true;
q.add(n);
}
}
}
//iterate over buckets and print each list
printMap(buckets);
}
//helper method that adds to Buckets list
private void addToBuckets(Map<Integer, List<Node>> buckets, Node n){
List<Node> currlist = buckets.get(n.height);
if (currlist==null)
{
List<Node> list = new ArrayList<Node>();
list.add(n);
buckets.put(n.height, list);
}
else{
currlist.add(n);
}
}
//prints the Map
private void printMap(Map<Integer, List<Node>> buckets){
for (Entry<Integer, List<Node>> e : buckets.entrySet()){
for (Node n : e.getValue()){
System.out.print(n.value + " ");
}
System.out.println();
}
Simplest way to do this without using any level information implicitly assumed to be in each Node. Just append a 'null' node after each level. check for this null node to know when to print a new line:
public class BST{
private Node<T> head;
BST(){}
public void setHead(Node<T> val){head = val;}
public static void printBinaryTreebyLevels(Node<T> head){
if(head == null) return;
Queue<Node<T>> q = new LinkedList<>();//assuming you have type inference (JDK 7)
q.add(head);
q.add(null);
while(q.size() > 0){
Node n = q.poll();
if(n == null){
System.out.println();
q.add(null);
n = q.poll();
}
System.out.print(n.value+" ");
if(n.left != null) q.add(n.left);
if(n.right != null) q.add(n.right);
}
}
public static void main(String[] args){
BST b = new BST();
c = buildListedList().getHead();//assume we have access to this for the sake of the example
b.setHead(c);
printBinaryTreeByLevels();
return;
}
}
class Node<T extends Number>{
public Node left, right;
public T value;
Node(T val){value = val;}
}
This works for me. Pass an array list with rootnode when calling printLevel.
void printLevel(ArrayList<Node> n){
ArrayList<Node> next = new ArrayList<Node>();
for (Node t: n) {
System.out.print(t.value+" ");
if (t.left!= null)
next.add(t.left);
if (t.right!=null)
next.add(t.right);
}
System.out.println();
if (next.size()!=0)
printLevel(next);
}
Print Binary Tree in level order with a single Queue:
public void printBFSWithQueue() {
java.util.LinkedList<Node> ll = new LinkedList<>();
ll.addLast(root);
ll.addLast(null);
Node in = null;
StringBuilder sb = new StringBuilder();
while(!ll.isEmpty()) {
if(ll.peekFirst() == null) {
if(ll.size() == 1) {
break;
}
ll.removeFirst();
System.out.println(sb);
sb = new StringBuilder();
ll.addLast(null);
continue;
}
in = ll.pollFirst();
sb.append(in.v).append(" ");
if(in.left != null) {
ll.addLast(in.left);
}
if(in.right != null) {
ll.addLast(in.right);
}
}
}
void printTreePerLevel(Node root)
{
Queue<Node> q= new LinkedList<Node>();
q.add(root);
int currentlevel=1;
int nextlevel=0;
List<Integer> values= new ArrayList<Integer>();
while(!q.isEmpty())
{
Node node = q.remove();
currentlevel--;
values.add(node.value);
if(node.left != null)
{
q.add(node.left);
nextlevel++;
}
if(node.right != null)
{
q.add(node.right);
nextlevel++;
}
if(currentlevel==0)
{
for(Integer i:values)
{
System.out.print(i + ",");
}
System.out.println();
values.clear();
currentlevel=nextlevel;
nextlevel=0;
}
}
}
Python implementation
# Function to print level order traversal of tree
def printLevelOrder(root):
h = height(root)
for i in range(1, h+1):
printGivenLevel(root, i)
# Print nodes at a given level
def printGivenLevel(root , level):
if root is None:
return
if level == 1:
print "%d" %(root.data),
elif level > 1 :
printGivenLevel(root.left , level-1)
printGivenLevel(root.right , level-1)
""" Compute the height of a tree--the number of nodes
along the longest path from the root node down to
the farthest leaf node
"""
def height(node):
if node is None:
return 0
else :
# Compute the height of each subtree
lheight = height(node.left)
rheight = height(node.right)
#Use the larger one
if lheight > rheight :
return lheight+1
else:
return rheight+1
Queue<Node> queue = new LinkedList<>();
queue.add(root);
Node leftMost = null;
while (!queue.isEmpty()) {
Node node = queue.poll();
if (leftMost == node) {
System.out.println();
leftMost = null;
}
System.out.print(node.getData() + " ");
Node left = node.getLeft();
if (left != null) {
queue.add(left);
if (leftMost == null) {
leftMost = left;
}
}
Node right = node.getRight();
if (right != null) {
queue.add(right);
if (leftMost == null) {
leftMost = right;
}
}
}
To solve this type of question which require in-level or same-level traversal approach, one immediately can use Breath First Search or in short BFS. To implement the BFS one can use Queue. In Queue each item comes in order of insertion, so for example if a node has two children, we can insert its children into queue one after another, thus make them in order inserted. When in return polling from queue, we traverse over children as it like we go in same-level of tree. Hense I am going to use a simple implementation of an in-order traversal approach.
I build up my Tree and pass the root which points to the root.
inorderTraversal takes root and do a while-loop that peeks one node first, and fetches children and insert them back into queue. Note that nodes one by one get inserted into queue, as you see, once you fetch the children nodes, you append it to the StringBuilder to construct the final output.
In levelOrderTraversal method though, I want to print the tree in level order. So I need to do the above approach, but instead I don't poll from queue and insert its children back to queue. Because I intent to insert "next-line-character" in a loop, and if I insert the children to queue, this loop would continue inserting a new line for each node, instead I need to check do it only for a level. That's why I used a for-loop to check how many items I have in my queue.
I simply don't poll anything from queue, because I only want to know if there are any level exists.
This separation of method helps me to still keep using BFS data and when required I can print them in-order or level-order , based-on requirements of the application.
public class LevelOrderTraversal {
public static void main(String[] args) throws InterruptedException {
BinaryTreeNode node1 = new BinaryTreeNode(100);
BinaryTreeNode node2 = new BinaryTreeNode(50);
BinaryTreeNode node3 = new BinaryTreeNode(200);
node1.left = node2;
node1.right = node3;
BinaryTreeNode node4 = new BinaryTreeNode(25);
BinaryTreeNode node5 = new BinaryTreeNode(75);
node2.left = node4;
node2.right = node5;
BinaryTreeNode node6 = new BinaryTreeNode(350);
node3.right = node6;
String levelOrderTraversal = levelOrderTraversal(node1);
System.out.println(levelOrderTraversal);
String inorderTraversal = inorderTraversal(node1);
System.out.println(inorderTraversal);
}
private static String inorderTraversal(BinaryTreeNode root) {
Queue<BinaryTreeNode> queue = new LinkedList<>();
StringBuilder sb = new StringBuilder();
queue.offer(root);
BinaryTreeNode node;
while ((node = queue.poll()) != null) {
sb.append(node.data).append(",");
if (node.left != null) {
queue.offer(node.left);
}
if (node.right != null) {
queue.offer(node.right);
}
}
return sb.toString();
}
public static String levelOrderTraversal(BinaryTreeNode root) {
Queue<BinaryTreeNode> queue = new LinkedList<>();
queue.offer(root);
StringBuilder stringBuilder = new StringBuilder();
while (!queue.isEmpty()) {
handleLevelPrinting(stringBuilder, queue);
}
return stringBuilder.toString();
}
private static void handleLevelPrinting(StringBuilder sb, Queue<BinaryTreeNode> queue) {
for (int i = 0; i < queue.size(); i++) {
BinaryTreeNode node = queue.poll();
if (node != null) {
sb.append(node.data).append("\t");
queue.offer(node.left);
queue.offer(node.right);
}
}
sb.append("\n");
}
private static class BinaryTreeNode {
int data;
BinaryTreeNode right;
BinaryTreeNode left;
public BinaryTreeNode(int data) {
this.data = data;
}
}
}
Wow. So many answers. For what it is worth, my solution goes like this:
We know the normal way to level order traversal: for each node, first the node is visited and then it’s child nodes are put in a FIFO queue. What we need to do is keep track of each level, so that all the nodes at that level are printed in one line, without a new line.
So I naturally thought of it as miaintaining a queue of queues. The main queue contains internal queues for each level. Each internal queue contains all the nodes in one level in FIFO order. When we dequeue an internal queue, we iterate through it, adding all its children to a new queue, and adding this queue to the main queue.
public static void printByLevel(Node root) {
Queue<Node> firstQ = new LinkedList<>();
firstQ.add(root);
Queue<Queue<Node>> mainQ = new LinkedList<>();
mainQ.add(firstQ);
while (!mainQ.isEmpty()) {
Queue<Node> levelQ = mainQ.remove();
Queue<Node> nextLevelQ = new LinkedList<>();
for (Node x : levelQ) {
System.out.print(x.key + " ");
if (x.left != null) nextLevelQ.add(x.left);
if (x.right != null) nextLevelQ.add(x.right);
}
if (!nextLevelQ.isEmpty()) mainQ.add(nextLevelQ);
System.out.println();
}
}
public void printAtLevel(int i){
printAtLevel(root,i);
}
private void printAtLevel(BTNode<T> n,int i){
if(n != null){
sop(n.data);
} else {
printAtLevel(n.left,i-1);
printAtLevel(n.right,i-1);
}
}
private void printAtLevel(BTNode<T> n,int i){
if(n != null){
sop(n.data);
printAtLevel(n.left,i-1);
printAtLevel(n.right,i-1);
}
}