Java Printing a Binary Tree using Level-Order in a Specific Format

Question

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. :)

Answer 1

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>();

        }

    }

Answer 2

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  

Answer 3

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)

Answer 4

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);
}

Answer 5

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;
}

Answer 6

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();
    }
}

Answer 7

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);
}

}

Answer 8

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());
}

Answer 9

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;
 }

Answer 10

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);
}

Answer 11

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);

}

Answer 12

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();
}

Answer 13

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;}
}

Answer 14

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);
}

Answer 15

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);
        }
    }
}

Answer 16

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;
            }


        }

    }

Answer 17

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 

Answer 18

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

Answer 19

        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;
                }
            }
        }

Answer 20

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;
    }
}

}

Answer 21

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.

1. First I calculate the height of the tree.
2. I then have a function to print a particular level of the tree.
3. 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.

Answer 22

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();
    }
}

Answer 23

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);
    }
}