Given singly Linked List: 1 -> 2 -> 3 -> 4 -> 5 -> null
Modify middle element as doubly Linked List Node
here middle element is 3
3 -> next should point to 4
3 -> prev should point to 1
Can any one suggest how can it be done ? interviewer gave me hint use interface. but I couldn't figure it out how.
I have to iterate over this linked list and print all the node and when it reaches to the middle, print where next and prev is pointing to, then print till the end of the list.
Expected output : 1, 2, Middle: 3, Prev: 1, Next: 4, 5
I'm facing problem in adding the middle node.
So, this "works", but if this is expected to be answered on an interview, it is way too much work.
LinkedList
public class LinkedList {
public interface Linkable<V, L extends Linkable> {
V getValue();
L getNext();
void setNext(L link);
}
public static class Node implements Linkable<Integer, Linkable> {
int value;
Linkable next;
Node(int value) {
this.value = value;
}
#Override
public Integer getValue() {
return value;
}
#Override
public Linkable getNext() {
return next;
}
#Override
public void setNext(Linkable link) {
this.next = link;
}
}
private Linkable head;
public boolean isEmpty() {
return this.head == null;
}
public Linkable getHead() {
return head;
}
public void add(int v) {
Node next = new Node(v);
if (isEmpty()) {
this.head = next;
} else {
Linkable tmp = this.head;
while (tmp.getNext() != null) {
tmp = tmp.getNext();
}
tmp.setNext(next);
}
}
}
Interface
interface DoublyLinkable<V, L extends LinkedList.Linkable> extends LinkedList.Linkable<V,L> {
LinkedList.Linkable getPrev();
void setPrev(LinkedList.Linkable prev);
}
DoubleNode
public class DoubleNode extends LinkedList.Node implements DoublyLinkable<Integer, LinkedList.Linkable> {
LinkedList.Linkable prev;
public DoubleNode(int value) {
super(value);
}
#Override
public LinkedList.Linkable getPrev() {
return prev;
}
#Override
public void setPrev(LinkedList.Linkable prev) {
this.prev = prev;
}
}
Driver
Outputs
1, 2, Middle: 3, Prev: 1, Next: 4, 5
public class Driver {
public static LinkedList getList() {
LinkedList list = new LinkedList();
for (int i = 1; i <= 5; i++) {
list.add(i);
}
return list;
}
public static void main(String[] args) {
LinkedList list = getList();
LinkedList.Linkable head = list.getHead();
LinkedList.Linkable beforeMiddle = null;
LinkedList.Linkable middle = list.getHead();
LinkedList.Linkable end = list.getHead();
if (head != null) {
// find the middle of the list
while (true) {
if (end.getNext() == null || end.getNext().getNext() == null) break;
beforeMiddle = middle;
middle = middle.getNext();
end = end.getNext().getNext();
}
// Replace middle by reassigning the pointer to it
if (beforeMiddle != null) {
DoubleNode n = new DoubleNode((int) middle.getValue()); // same value
n.setPrev(list.getHead()); // point back to the front
n.setNext(middle.getNext()); // point forward to original value
beforeMiddle.setNext((DoublyLinkable) n);
middle = beforeMiddle.getNext();
}
// Build the "expected" output
StringBuilder sb = new StringBuilder();
final String DELIMITER = ", ";
head = list.getHead();
boolean atMiddle = false;
if (head != null) {
do {
if (head instanceof DoublyLinkable) {
atMiddle = true;
String out = String.format("Middle: %d, Prev: %d, ", (int) head.getValue(), (int) ((DoublyLinkable) head).getPrev().getValue());
sb.append(out);
} else {
if (atMiddle) {
sb.append("Next: ");
atMiddle = false;
}
sb.append(head.getValue()).append(DELIMITER);
}
head = head.getNext();
} while (head != null);
}
sb.setLength(sb.length() - DELIMITER.length());
System.out.println(sb.toString());
}
}
}
By definition, a single-linked list consists of single-linked nodes only, and a double-linked consists of double-linked nodes only. Otherwise. it is neither.
By definition the field prev of a double-linked list must point to the previous element.
Whatever you are supposed to build. It's something not well specified. So if you really were asked this in an interview (and did not misunderstand the question - maybe he wanted you to point out that ghis violates the interface?) this is a case for the code horror stories of http://thedailywtf.com/ - section "incompetent interviewers".
If you haven't, you'd better define a lenght() function so given one linked list you can know how many nodes does it have.
Thanks to the response of Cereal_Killer to the previous version of this answer, I noticed that the list is firstly a singly linked list, and you just have to make the middle node be linked both to the next node and to some previous node.
Now I guess that you have defined two structures (Struct, Class or whatever depending on the language you're using). So lets say you have Node_s defined as a node with only a next pointer, and Node_d with both a next and a prev pointer. (Node_d may inherite from Node_s so you just have to add the prev attribute in the child class). Knowing this, the code above should be doing what you need:
function do_it(my_LinkedList linkedList){
int i_middle;
int length = linkedList.length();
if ( (length ÷ 2 ) != 0 ) i_middle = length / 2;
else return -1;
Node_s aux = linkedList.first();
int index = 0;
Node_d middle= null;
while (aux != null) {
if (index == i_middle - 1){ //now aux is the middle's previous node
middle.data = aux.next.data; //aux.next is the middle singly node, we assignate the data to the new double linked node
middle.prev = aux; //as we said, aux is the prev to the middle
midle.next = aux.next.next; //so aux.next.next is the next to the middle
print(what you need to print);
}else {
print("Node " + index " next: "+ aux.next);
}//end if
index++;
aux = aux.next;
} //end while
}//end function
This previous code should be doing what you need. I wrote the answer in some kind of pseudo-java code so if you're not familiar with Java or don't understand what my pseudo-code does, please let me know. Anyway, the idea of my code may present some troubles depending on the language you're working with, so you'll have to adapt it.
Note that at the end of the execution of this program, your data structure won't be a singly linked list, and neither a double one, since you'll have linkedList.length() - 1 nodes linked in a signly way but the middle one will have two links.
Hope this helps.
I've been working through some standard coding interview questions from a book I recently bought, and I came across the following question and answer:
Implement an algorithm to find the nth to last element in a linked list.
Here's the provided answer:
public static LinkedListNode findNtoLast(LinkedListNode head, int n) { //changing LinkedListNode to ListNode<String>
if(head == null || n < 1) {
return null;
}
LinkedListNode p1 = head;
LinkedListNode p2 = head;
for(int j = 0; j < n-1; ++j) {
if(p2 == null) {
return null;
}
p2 = p2.next;
}
if(p2 == null) {
return null;
}
while(p2.next != null) {
p1 = p1.next;
p2 = p2.next;
}
return p1;
}
I understand the algorithm, how it works, and why the book lists this as its answer, but I'm confused about how to access the LinkedListNodes to send as an argument to the method. I know that I'd have to create a LinkedListNode class (since Java doesn't already have one), but I can't seem to figure out how to do that. It's frustrating because I feel like I should know how to do this. Here's something that I've been working on. I'd greatly appreciate any clarification. You can expand/comment on my code or offer your own alternatives. Thanks.
class ListNode<E> {
ListNode<E> next;
E data;
public ListNode(E value) {
data = value;
next = null;
}
public ListNode(E value, ListNode<E> n) {
data = value;
next = n;
}
public void setNext(ListNode<E> n) {
next = n;
}
}
public class MyLinkedList<E> extends LinkedList {
LinkedList<ListNode<E>> list;
ListNode<E> head;
ListNode<E> tail;
ListNode<E> current;
ListNode<E> prev;
public MyLinkedList() {
list = null;
head = null;
tail = null;
current = null;
prev = null;
}
public MyLinkedList(LinkedList<E> paramList) {
list = (LinkedList<ListNode<E>>) paramList; //or maybe create a loop assigning each ListNode a value and next ptr
head = list.getFirst();
tail = list.getLast(); //will need to update tail every time add new node
current = null;
prev = null;
}
public void addNode(E value) {
super.add(value);
//ListNode<E> temp = tail;
current = new ListNode<E>(value);
tail.setNext(current);
tail = current;
}
public LinkedList<ListNode<E>> getList() {
return list;
}
public ListNode<E> getHead() {
return head;
}
public ListNode<E> getTail() {
return tail;
}
public ListNode<E> getCurrent() {
return current;
}
public ListNode<E> getPrev() {
return prev;
}
}
How can the LinkedListNode head from a LinkedList?
Update: I think part of my confusion comes from what to put in the main method. Do I need to create a LinkedList of ListNode? If I do that, how would I connect the ListNodes to each other? How would I connect them without using a LinkedList collection object? If someone could show me how they would code the main method, I think that would put things into enough perspective for me to solve my issues. Here's my latest attempt at the main method:
public static void main(String args[]) {
LinkedList<ListNode<String>> list = new LinkedList<ListNode<String>>();
//MyLinkedList<ListNode<String>> list = new MyLinkedList(linkedList);
list.add(new ListNode<String>("Jeff"));
list.add(new ListNode<String>("Brian"));
list.add(new ListNode<String>("Negin"));
list.add(new ListNode<String>("Alex"));
list.add(new ListNode<String>("Alaina"));
int n = 3;
//ListIterator<String> itr1 = list.listIterator();
//ListIterator<String> itr2 = list.listIterator();
LinkedListNode<String> head = new LinkedListNode(list.getFirst(), null);
//String result = findNtoLast(itr1, itr2, n);
//System.out.println("The " + n + "th to the last value: " + result);
//LinkedListNode<String> nth = findNtoLast(list.getFirst(), n);
ListNode<String> nth = findNtoLast(list.getFirst(), n);
System.out.println("The " + n + "th to the last value: " + nth);
}
In an attempt to connect the nodes without using a custom linked list class, I have edited my ListNode class to the following:
class ListNode<E> {
ListNode<E> next;
ListNode<E> prev; //only used for linking nodes in singly linked list
ListNode<E> current; //also only used for linking nodes in singly linked list
E data;
private static int size = 0;
public ListNode() {
data = null;
next = null;
current = null;
if(size > 0) { //changed from prev != null because no code to make prev not null
prev.setNext(this);
}
size++;
}
public ListNode(E value) {
data = value;
next = null;
current = this;
System.out.println("current is " + current);
if(size > 0) {
prev.setNext(current);//this line causing npe
}
else
{
prev = current;
System.out.println("prev now set to " + prev);
}
size++;
System.out.println("after constructor, size is " + size);
}
public ListNode(E value, ListNode<E> n) {
data = value;
next = n;
current = this;
if(size > 0) {
prev.setNext(this);
}
size++;
}
public void setNext(ListNode<E> n) {
next = n;
}
}
As is right now, the program will run until it reaches prev.setNext(current); in the single argument constructor for ListNode. Neither current nor prev are null at the time this line is reached. Any advice would be greatly appreciated. Thanks.
You don't actually need a separate LinkedList class; the ListNode class is a linked list. Or, to state it differently, a reference to the head of the list is a reference to the list.
The use of head, tail, current, prev in the sample code you posted has come from a double-linked list which is a data type that has links in both directions. This is more efficient for certain types of applications (such as finding the nth last item).
So I would recommend renaming your ListNode class to LinkedList and renaming next to tail.
To add a new item to the list you need a method that creates a new list with the new item at it's head. Here is an example:
class LinkedList<E> {
...
private LinkedList(E value, LinkedList<E> tail) {
this.data = value;
this.tail = tail;
}
public LinkedList<E> prependItem(E item) {
return new LinkedList(item, this);
}
}
Then to add a new item i to list you use list = list.prependItem(i);
If for some reason you need to always add the items to the end, then:
private LinkedList(E value) {
this.data = value;
this.tail = null;
}
public void appendItem(E item) {
LinkedList<E> list = this;
while (list.tail != null)
list = list.tail;
list.tail = new LinkedList<>(item);
}
However this is obviously pretty inefficient for long lists. If you need to do this then either use a different data structure or just reverse the list when you have finished adding to it.
Incidentally, an interesting side effect of this is that a reference to any item in the list is a reference to a linked list. This makes recursion very easy. For example, here's a recursive solution for finding the length of a list:
public int getLength(LinkedList list) {
if (list == null) {
return 0;
} else {
return 1 + getLength(list.getTail());
}
}
And using this a simple (but very inefficient!) solution to the problem you provided - I've renamed the method to make its function more obvious:
public LinkedList getTailOfListOfLengthN(LinkedList list, int n) {
int length = getLength(list);
if (length < n) {
return null;
} else if (length == n) {
return list;
} else {
return getTailOfLengthN(list.getTail(), n);
}
}
And to reverse the list:
public LinkedList<E> reverse() {
if (tail == null) {
return this;
} else {
LinkedList<E> list = reverse(tail);
tail.tail = this;
tail = null;
return list;
}
}
As I hope you can see this makes the methods a lot more elegant than separating the node list classes.
Actually you have created a linked list with you class ListNode.
A linked list is made of a node and a reference to another linked list (see the recursion?).
So I've been trying to sort a linked list using merge sort, I found this code and tried to work on it, but it doesn't seen to really work?
What could be the problem with it? I'm not quite sure about the getMiddle method although I know it should get the middle value of the list in order to work on 2 lists from the list itself
Here's the code;
public Node mergeSort(Node head) {
if (head == null || head.link == null) {
return head;
}
Node middle = getMiddle(head);
Node sHalf = middle.link;
middle.link = null;
return merge(mergeSort(head), mergeSort(sHalf));
}
public Node merge(Node a, Node b) {
Node dummyHead;
Node current;
dummyHead = new Node();
current = dummyHead;
while (a != null && b != null) {
if ((int) a.getData() <= (int) b.getData()) {
current.link = a;
a.link = a;
}
else {
current.link = b;
b.link = a;
}
current = current.link;
}
current.link = (a == null) ? b : a;
return dummyHead;
}
public Node getMiddle(Node head) {
if (head == null) {
return head;
}
Node slow, fast;
slow = fast = head;
while (fast.link != null && fast.link.link != null) {
slow = slow.link;
fast = fast.link.link;
}
return slow;
}
In the main method:
Object data;
MyLinkedList list = new MyLinkedList(); //empty list.
for (int i = 0; i < 3; i++) { //filling the list
data = console.nextInt();
list.insertAtFront(data);
}
System.out.print("Print(1): ");
list.printList();
list.mergeSort(list.getHead());
System.out.print("List after sorting: ");
list.printList();
One problem is the getMiddle method doesn't correctly return the middle Node.
Consider a linked list with 5 Nodes (a, b, c, d, e)
head, slow, and fast all begin at index 0 (a).
After the first iteration of the while loop, slow is at 1 (b) and fast is at 2 (c); after the second iteration, slow is at 2 (c) and fast at 4 (e). These are both not null, so another iteration happens, putting slow at at 3 (d) and fast at null. Since fast is null, the while loop is exited and slow is returned; however slow has node 3 (d) rather than the middle node, 2 (c).
An alternate way to get the middle node would be to simply use the number of nodes:
public Node getMiddle(Node head) {
Node counter = head;
int numNodes = 0;
while(counter != null) {
counter = counter.link;
numNodes++;
}
if(numNodes == 0)
return null;
Node middle = head;
for(int i=0; i<numNodes/2; i++)
middle = middle.link;
return middle;
}
I your mergeSort method is fine, but technically it only needs to return head if head.link is null, not if head itself is null (since that would never happen anyhow):
public Node mergeSort(Node head) {
if (head.link == null) {
return head;
}
// same
}
Most importantly, your merge method. You can write a public void setData(Object) method in your Node class to make this easier. The following code should work, although I can't claim it's the best/most efficient way to do the job
public Node merge(Node a, Node b) {
Node combined = new Node();
Node current = combined;
while(a != null || b != null) {
if(a == null)
addNode(current, b);
if(b == null)
addNode(current, a);
if((int)a.getData()<(int)b.getData())
addNode(current, a);
else
addNode(current, b);
}
return combined;
}
Uses the following helper method:
public void addNode(Node n1, Node n2) {
n1.setData((int)n2.getData());
n1.link = new Node();
n1 = n1.link;
n2 = n2.link
}
I am currently revising for my programming exam and I have came across a question from a past paper that has me rather confused.
I have two classes, Queue and Node, shown below.
The question states that I have to extend the behaviour of the Queue class by adding the necessary code to the inspectQueue method that prints to the console all the data stored within the queue.
The only solution I can think of, and it is very weak, is to have a simple ArrayList and every time an element is enqueued/dequeued then add/remove the node to/from the list.
Is there a better solution that I am glossing over? I'd really appreciate some guidance.
I've commented the code where I have implemented my "solution" the rest of the code is how it appears in the exam paper.
Thanks for your time.
Queue.java
public class Queue {
protected Node head;
protected Node last;
//added by me
private ArrayList<Node> nodes = new ArrayList<Node>();
//end my add
public boolean isEmpty() {
return (this.head == null);
}
public void enqueue(Object d) {
Node n = new Node();
n.setData(d);
nodes.add(n); //added by me
if (this.isEmpty()) {
head = n;
last = n;
}
else {
last.setNext(n);
last = n;
}
}
public Object dequeue() {
if(this.isEmpty()) {
this.last = null;
return null;
}
else {
Node h = this.head;
nodes.remove(h); //added by me
head = h.getNext();
return h.getData();
}
}
public Object peek() {
if(this.isEmpty()) {
return null;
}
else {
Node t = this.head;
return t.getData();
}
}
public void clearQueue() {
this.head = null;
this.last = null;
}
public void inspectQueue() {
//added by me (all below)
System.out.println("Inspecting Queue: (contains " + nodes.size() + " nodes)");
for(Node n : nodes) {
System.out.println(n.getData());
}
}
}
Node.java
public class Node {
protected Object data;
protected Node next;
public void setNext(Node e) {
this.next = e;
}
public Node getNext() {
return this.next;
}
public void setData(Object d) {
this.data = d;
}
public Object getData() {
return this.data;
}
}
Your nodes form a linked list, so just do
public void inspectQueue() {
Node n = head;
while (n != null) {
System.out.println(n.getData());
n = n.getNext();
}
}
This is a very basic data structure, called a LinkedList. In the your code for the Node class you can see the following:
protected Node next;
This means that every Node also holds a reference to the next Node in the list. If this Node is null, there are no more elements in the list. Knowing this, you can loop somewhat like this:
Node currentNode = this.head;
while(currentNode != null) {
System.out.println(currentNode.getData().toString());
currentNode = currentNode.getNext();
}
This eliminates the need for an ArrayList to store your references.
The LinkedList is a VERY frequently used data structure and very important to understand. If you have any questions, just go ahead and ask!
If you also want to have the size, keep a counter along, increment it each time you call getNext(), and print the size after the for loop.
You don't need the array, you have that information stored within the Node next property:
public void inspectQueue() {
Node current = head;
while(current != null) {
System.out.println(n.getData());
current = current.getNext();
}
}
That data structure is called linked list.
The simpler solution is to start with queue.head and traverse the linked list of nodes using node.next, printing the data as you go along.
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);
}
}