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Print level order traversal line by line | Set 1

  • Difficulty Level : Easy
  • Last Updated : 01 Jul, 2021
 

Given a binary tree, print level order traversal in a way that nodes of all levels are printed in separate lines.
For example consider the following tree 
 

Example 1:

Output for above tree should be
20
8 22
4 12
10 14

Example 2:
          1
       /     \
      2       3
    /   \       \
   4     5       6
        /  \     /
       7    8   9
Output for above tree should be
1
2 3
4 5 6
7 8 9<
 

Note that this is different from simple level order traversal where we need to print all nodes together. Here we need to print nodes of different levels in different lines.
A simple solution is to print use the recursive function discussed in the level order traversal post and print a new line after every call to printGivenLevel().
 

C++




/* Function to line by line print level order traversal a tree*/
void printLevelOrder(struct node* root)
{
    int h = height(root);
    int i;
    for (i=1; i<=h; i++)
    {
        printGivenLevel(root, i);
        printf("\n");
    }
}
 
/* Print nodes at a given level */
void printGivenLevel(struct node* root, int level)
{
    if (root == NULL)
        return;
    if (level == 1)
        printf("%d ", root->data);
    else if (level > 1)
    {
        printGivenLevel(root->left, level-1);
        printGivenLevel(root->right, level-1);
    }
}

Java




/* Function to line by line print level order traversal a tree*/
static void printLevelOrder(Node root)
{
    int h = height(root);
    int i;
    for (i=1; i<=h; i++)
    {
        printGivenLevel(root, i);
        System.out.println();
    }
}
/* Print nodes at a given level */
void printGivenLevel(Node root, int level)
{
    if (root == null)
        return;
    if (level == 1)
        System.out.println(root.data);
    else if (level > 1)
    {
        printGivenLevel(root.left, level-1);
        printGivenLevel(root.right, level-1);
    }
}

Python3




# Python3 program for above approach
 
def printlevelorder(root):
    h = height(root)
    for i in range(1, h + 1):
        givenspirallevel(root, i)
 
def printGivenLevel(root, level):
    if root is None:
        return root
     
    if level == 1:
        print(root.val, end = ' ')
    elif level > 1:
        printGivenLevel(root.left, level - 1)
        printGivenLevel(root.right, level - 1)
 
# This code is contributed by Praveen kumar

C#




/* Print nodes at a given level */
static void printGivenLevel(Node root, int level)
{
    if (root == null)
        return;
    if (level == 1)
        Console.WriteLine(root.data);
    else if (level > 1)
    {
        printGivenLevel(root.left, level-1);
        printGivenLevel(root.right, level-1);
    }
}

Javascript




/* Print nodes at a given level */
function printGivenLevel(root, level)
{
    if (root == null)
        return;
    if (level == 1)
        document.write(root.data);
    else if (level > 1)
    {
        printGivenLevel(root.left, level-1);
        printGivenLevel(root.right, level-1);
    }
}

The time complexity of the above solution is O(n2)
How to modify the iterative level order traversal (Method 2 of this) to levels line by line? 
The idea is similar to this post. We count the nodes at current level. And for every node, we enqueue its children to queue. 
 

C++




/* Iterative program to print levels line by line */
#include <iostream>
#include <queue>
using namespace std;
 
// A Binary Tree Node
struct node
{
    struct node *left;
    int data;
    struct node *right;
};
 
// Iterative method to do level order traversal
// line by line
void printLevelOrder(node *root)
{
    // Base Case
    if (root == NULL) return;
 
    // Create an empty queue for level order traversal
    queue<node *> q;
 
    // Enqueue Root and initialize height
    q.push(root);
 
    while (q.empty() == false)
    {
        // nodeCount (queue size) indicates number
        // of nodes at current level.
        int nodeCount = q.size();
 
        // Dequeue all nodes of current level and
        // Enqueue all nodes of next level
        while (nodeCount > 0)
        {
            node *node = q.front();
            cout << node->data << " ";
            q.pop();
            if (node->left != NULL)
                q.push(node->left);
            if (node->right != NULL)
                q.push(node->right);
            nodeCount--;
        }
        cout << endl;
    }
}
 
// Utility function to create a new tree node
node* newNode(int data)
{
    node *temp = new node;
    temp->data = data;
    temp->left = NULL;
    temp->right = NULL;
    return temp;
}
 
// Driver program to test above functions
int main()
{
    // Let us create binary tree shown above
    node *root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);
    root->right->right = newNode(6);
 
    printLevelOrder(root);
    return 0;
}

Java




/* An Iterative Java program to print levels line by line */
 
import java.util.LinkedList;
import java.util.Queue;
 
public class LevelOrder
{
    // A Binary Tree Node
    static class Node
    {
        int data;
        Node left;
        Node right;
         
        // constructor
        Node(int data){
            this.data = data;
            left = null;
            right =null;
        }
    }
     
    // Iterative method to do level order traversal line by line
    static void printLevelOrder(Node root)
    {
        // Base Case
        if(root == null)
            return;
         
        // Create an empty queue for level order traversal
        Queue<Node> q =new LinkedList<Node>();
         
        // Enqueue Root and initialize height
        q.add(root);
         
         
        while(true)
        {
             
            // nodeCount (queue size) indicates number of nodes
            // at current level.
            int nodeCount = q.size();
            if(nodeCount == 0)
                break;
             
            // Dequeue all nodes of current level and Enqueue all
            // nodes of next level
            while(nodeCount > 0)
            {
                Node node = q.peek();
                System.out.print(node.data + " ");
                q.remove();
                if(node.left != null)
                    q.add(node.left);
                if(node.right != null)
                    q.add(node.right);
                nodeCount--;
            }
            System.out.println();
        }
    }
     
    // Driver program to test above functions
    public static void main(String[] args)
    {
        // Let us create binary tree shown in above diagram
       /*               1
                   /     \
                  2       3
                /   \       \
               4     5       6
        */
         
        Node root = new Node(1);
        root.left = new Node(2);
        root.right = new Node(3);
        root.left.left = new Node(4);
        root.left.right = new Node(5);
        root.right.right = new Node(6);
         
        printLevelOrder(root);
 
    }
 
}
//This code is contributed by Sumit Ghosh

Python3




# Python3 program for above approach
class newNode:
    def __init__(self, data):
        self.val = data
        self.left = None
        self.right = None
         
# Iterative method to do level order traversal
# line by line
def printLevelOrder(root):
     
    # Base case
    if root is None:
        return
    # Create an empty queue for level order traversal
    q = []
     
    # Enqueue root and initialize height
    q.append(root)
         
    while q:
     
        # nodeCount (queue size) indicates number
        # of nodes at current level.
        count = len(q)
         
        # Dequeue all nodes of current level and
        # Enqueue all nodes of next level
        while count > 0:
            temp = q.pop(0)
            print(temp.val, end = ' ')
            if temp.left:
                q.append(temp.left)
            if temp.right:
                q.append(temp.right)
 
            count -= 1
        print(' ')
         
# Driver Code
root = newNode(1);
root.left = newNode(2);
root.right = newNode(3);
root.left.left = newNode(4);
root.left.right = newNode(5);
root.right.right = newNode(6);
 
printLevelOrder(root);
 
# This code is contributed by Praveen kumar

C#




/* An Iterative C# program to print
levels line by line */
using System;
using System.Collections.Generic;
 
public class LevelOrder
{
    // A Binary Tree Node
    class Node
    {
        public int data;
        public Node left;
        public Node right;
         
        // constructor
        public Node(int data)
        {
            this.data = data;
            left = null;
            right =null;
        }
    }
     
    // Iterative method to do level order
    // traversal line by line
    static void printLevelOrder(Node root)
    {
        // Base Case
        if(root == null)
            return;
         
        // Create an empty queue for level
        // order traversal
        Queue<Node> q =new Queue<Node>();
         
        // Enqueue Root and initialize height
        q.Enqueue(root);
         
        while(true)
        {
             
            // nodeCount (queue size) indicates
            // number of nodes at current level.
            int nodeCount = q.Count;
            if(nodeCount == 0)
                break;
             
            // Dequeue all nodes of current level
            // and Enqueue all nodes of next level
            while(nodeCount > 0)
            {
                Node node = q.Peek();
                Console.Write(node.data + " ");
                q.Dequeue();
                if(node.left != null)
                    q.Enqueue(node.left);
                if(node.right != null)
                    q.Enqueue(node.right);
                nodeCount--;
            }
            Console.WriteLine();
        }
    }
     
    // Driver Code
    public static void Main(String[] args)
    {
        // Let us create binary tree shown
        // in above diagram
        /*         1
                    / \
                    2 3
                    / \ \
                4 5 6
            */
        Node root = new Node(1);
        root.left = new Node(2);
        root.right = new Node(3);
        root.left.left = new Node(4);
        root.left.right = new Node(5);
        root.right.right = new Node(6);
         
        printLevelOrder(root);
    }
}
 
// This code is contributed 29AjayKumar

Javascript




<script>
 
    /* An Iterative Javascript program to
       print levels line by line */
     
    class Node
    {
        constructor(data) {
           this.left = null;
           this.right = null;
           this.data = data;
        }
    }
     
    // Iterative method to do level order traversal line by line
    function printLevelOrder(root)
    {
        // Base Case
        if(root == null)
            return;
          
        // Create an empty queue for level order traversal
        let q = [];
          
        // Enqueue Root and initialize height
        q.push(root);
          
          
        while(true)
        {
              
            // nodeCount (queue size) indicates number of nodes
            // at current level.
            let nodeCount = q.length;
            if(nodeCount == 0)
                break;
              
            // Dequeue all nodes of current level and Enqueue all
            // nodes of next level
            while(nodeCount > 0)
            {
                let node = q[0];
                document.write(node.data + " ");
                q.shift();
                if(node.left != null)
                    q.push(node.left);
                if(node.right != null)
                    q.push(node.right);
                nodeCount--;
            }
            document.write("</br>");
        }
    }
     
    // Let us create binary tree shown in above diagram
    /*                  1
                     /     \
                    2       3
                  /   \       \
                 4     5       6
          */
 
    let root = new Node(1);
    root.left = new Node(2);
    root.right = new Node(3);
    root.left.left = new Node(4);
    root.left.right = new Node(5);
    root.right.right = new Node(6);
 
    printLevelOrder(root);
     
</script>

Output:  



1
2 3
4 5 6

 

Time complexity of this method is O(n) where n is number of nodes in given binary tree.
Level order traversal line by line | Set 2 (Using Two Queues)
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
 

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