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Level Order Binary Tree Traversal

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  • Difficulty Level : Easy
  • Last Updated : 15 Nov, 2022
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Given the root of the Binary Tree. The task is to print the Level order traversal of a tree is breadth first traversal for the tree. 

Input:

Output:
1
2 3
4 5 

Input:

Output:
20
8 22
4 12
10 14

Recommended Practice

Level Order Binary Tree Traversal using Recursion:

Below is the idea to solve the problem:

Print the level order traversal of the tree using recursive function to traverse all nodes of a level. Find height of tree and run depth first search and maintain current height, print nodes for every height from root and for 1 to height and match if the current height is equal to height of the iteration then print node’s data.

Follow the below steps to Implement the idea:

  • Run a for loop for counter i, i.e. current height from 1 to h (height of the tree).
  • Use DFS to traverse the tree and maintain height for the current node.
    • If the Node is NULL then return;
    • If level is 1 print(tree->data);
    • Else if the level is greater than 1, then
      • Recursively call to for tree->left, level-1.
      • Recursively call to for tree->right, level-1.

Below is the implementation of the above approach:

C++




// Recursive CPP program for level
// order traversal of Binary Tree
#include <bits/stdc++.h>
using namespace std;
 
/* A binary tree node has data,
pointer to left child
and a pointer to right child */
class node {
public:
    int data;
    node *left, *right;
};
 
/* Function prototypes */
void printCurrentLevel(node* root, int level);
int height(node* node);
node* newNode(int data);
 
/* Function to print level
order traversal a tree*/
void printLevelOrder(node* root)
{
    int h = height(root);
    int i;
    for (i = 1; i <= h; i++)
        printCurrentLevel(root, i);
}
 
/* Print nodes at a current level */
void printCurrentLevel(node* root, int level)
{
    if (root == NULL)
        return;
    if (level == 1)
        cout << root->data << " ";
    else if (level > 1) {
        printCurrentLevel(root->left, level - 1);
        printCurrentLevel(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.*/
int height(node* node)
{
    if (node == NULL)
        return 0;
    else {
        /* compute the height of each subtree */
        int lheight = height(node->left);
        int rheight = height(node->right);
 
        /* use the larger one */
        if (lheight > rheight) {
            return (lheight + 1);
        }
        else {
            return (rheight + 1);
        }
    }
}
 
/* Helper function that allocates
a new node with the given data and
NULL left and right pointers. */
node* newNode(int data)
{
    node* Node = new node();
    Node->data = data;
    Node->left = NULL;
    Node->right = NULL;
 
    return (Node);
}
 
/* Driver code*/
int main()
{
    node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);
 
    cout << "Level Order traversal of binary tree is \n";
    printLevelOrder(root);
 
    return 0;
}
 
// This code is contributed by rathbhupendra

C




// Recursive C program for level
// order traversal of Binary Tree
#include <stdio.h>
#include <stdlib.h>
 
/* A binary tree node has data,
   pointer to left child
   and a pointer to right child */
struct node {
    int data;
    struct node *left, *right;
};
 
/* Function prototypes */
void printCurrentLevel(struct node* root, int level);
int height(struct node* node);
struct node* newNode(int data);
 
/* Function to print level order traversal a tree*/
void printLevelOrder(struct node* root)
{
    int h = height(root);
    int i;
    for (i = 1; i <= h; i++)
        printCurrentLevel(root, i);
}
 
/* Print nodes at a current level */
void printCurrentLevel(struct node* root, int level)
{
    if (root == NULL)
        return;
    if (level == 1)
        printf("%d ", root->data);
    else if (level > 1) {
        printCurrentLevel(root->left, level - 1);
        printCurrentLevel(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.*/
int height(struct node* node)
{
    if (node == NULL)
        return 0;
    else {
        /* compute the height of each subtree */
        int lheight = height(node->left);
        int rheight = height(node->right);
 
        /* use the larger one */
        if (lheight > rheight)
            return (lheight + 1);
        else
            return (rheight + 1);
    }
}
 
/* Helper function that allocates a new node with the
   given data and NULL left and right pointers. */
struct node* newNode(int data)
{
    struct node* node
        = (struct node*)malloc(sizeof(struct node));
    node->data = data;
    node->left = NULL;
    node->right = NULL;
 
    return (node);
}
 
/* Driver program to test above functions*/
int main()
{
    struct node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);
 
    printf("Level Order traversal of binary tree is \n");
    printLevelOrder(root);
 
    return 0;
}

Java




// Recursive Java program for level
// order traversal of Binary Tree
 
/* Class containing left and right child of current
   node and key value*/
class Node {
    int data;
    Node left, right;
    public Node(int item)
    {
        data = item;
        left = right = null;
    }
}
 
class BinaryTree {
    // Root of the Binary Tree
    Node root;
 
    public BinaryTree() { root = null; }
 
    /* function to print level order traversal of tree*/
    void printLevelOrder()
    {
        int h = height(root);
        int i;
        for (i = 1; i <= h; i++)
            printCurrentLevel(root, i);
    }
 
    /* Compute the "height" of a tree -- the number of
    nodes along the longest path from the root node
    down to the farthest leaf node.*/
    int height(Node root)
    {
        if (root == null)
            return 0;
        else {
            /* compute  height of each subtree */
            int lheight = height(root.left);
            int rheight = height(root.right);
 
            /* use the larger one */
            if (lheight > rheight)
                return (lheight + 1);
            else
                return (rheight + 1);
        }
    }
 
    /* Print nodes at the current level */
    void printCurrentLevel(Node root, int level)
    {
        if (root == null)
            return;
        if (level == 1)
            System.out.print(root.data + " ");
        else if (level > 1) {
            printCurrentLevel(root.left, level - 1);
            printCurrentLevel(root.right, level - 1);
        }
    }
 
    /* Driver program to test above functions */
    public static void main(String args[])
    {
        BinaryTree tree = new BinaryTree();
        tree.root = new Node(1);
        tree.root.left = new Node(2);
        tree.root.right = new Node(3);
        tree.root.left.left = new Node(4);
        tree.root.left.right = new Node(5);
 
        System.out.println("Level order traversal of"
                           + "binary tree is ");
        tree.printLevelOrder();
    }
}

Python3




# Recursive Python program for level
# order traversal of Binary Tree
 
# A node structure
 
 
class Node:
 
    # A utility function to create a new node
    def __init__(self, key):
        self.data = key
        self.left = None
        self.right = None
 
 
# Function to  print level order traversal of tree
def printLevelOrder(root):
    h = height(root)
    for i in range(1, h+1):
        printCurrentLevel(root, i)
 
 
# Print nodes at a current level
def printCurrentLevel(root, level):
    if root is None:
        return
    if level == 1:
        print(root.data, end=" ")
    elif level > 1:
        printCurrentLevel(root.left, level-1)
        printCurrentLevel(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
 
 
# Driver program to test above function
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.left.right = Node(5)
 
print("Level order traversal of binary tree is -")
printLevelOrder(root)
 
# This code is contributed by Nikhil Kumar Singh(nickzuck_007)

C#




// Recursive c# program for level
// order traversal of Binary Tree
using System;
 
/* Class containing left and right
   child of current node and key value*/
public class Node {
    public int data;
    public Node left, right;
    public Node(int item)
    {
        data = item;
        left = right = null;
    }
}
 
class GFG {
    // Root of the Binary Tree
    public Node root;
 
    public void BinaryTree() { root = null; }
 
    /* function to print level order
       traversal of tree*/
    public virtual void printLevelOrder()
    {
        int h = height(root);
        int i;
        for (i = 1; i <= h; i++) {
            printCurrentLevel(root, i);
        }
    }
 
    /* Compute the "height" of a tree --
    the number of nodes along the longest
    path from the root node down to the
    farthest leaf node.*/
    public virtual int height(Node root)
    {
        if (root == null) {
            return 0;
        }
        else {
            /* compute height of each subtree */
            int lheight = height(root.left);
            int rheight = height(root.right);
 
            /* use the larger one */
            if (lheight > rheight) {
                return (lheight + 1);
            }
            else {
                return (rheight + 1);
            }
        }
    }
 
    /* Print nodes at the current level */
    public virtual void printCurrentLevel(Node root,
                                          int level)
    {
        if (root == null) {
            return;
        }
        if (level == 1) {
            Console.Write(root.data + " ");
        }
        else if (level > 1) {
            printCurrentLevel(root.left, level - 1);
            printCurrentLevel(root.right, level - 1);
        }
    }
 
    // Driver Code
    public static void Main(string[] args)
    {
        GFG tree = new GFG();
        tree.root = new Node(1);
        tree.root.left = new Node(2);
        tree.root.right = new Node(3);
        tree.root.left.left = new Node(4);
        tree.root.left.right = new Node(5);
 
        Console.WriteLine("Level order traversal "
                          + "of binary tree is ");
        tree.printLevelOrder();
    }
}
 
// This code is contributed by Shrikant13

Javascript




<script>
// Recursive javascript program for level
// order traversal of Binary Tree
 
/* Class containing left and right child of current
   node and key value*/
 class Node {
        constructor(val) {
            this.data = val;
            this.left = null;
            this.right = null;
        }
    }
 
 
    // Root of the Binary Tree
    var root= null;
     
    /* function to print level order traversal of tree */
    function printLevelOrder() {
        var h = height(root);
        var i;
        for (i = 1; i <= h; i++)
            printCurrentLevel(root, i);
    }
 
    /*
     * Compute the "height" of a tree -- the number of nodes along the longest path
     * from the root node down to the farthest leaf node.
     */
    function height(root) {
        if (root == null)
            return 0;
        else {
            /* compute height of each subtree */
            var lheight = height(root.left);
            var rheight = height(root.right);
 
            /* use the larger one */
            if (lheight > rheight)
                return (lheight + 1);
            else
                return (rheight + 1);
        }
    }
 
    /* Print nodes at the current level */
    function printCurrentLevel(root , level) {
        if (root == null)
            return;
        if (level == 1)
            document.write(root.data + " ");
        else if (level > 1) {
            printCurrentLevel(root.left, level - 1);
            printCurrentLevel(root.right, level - 1);
        }
    }
 
    /* Driver program to test above functions */
     
        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);
 
       document.write("Level order traversal of  binary tree is ");
       printLevelOrder();
 
// This code is contributed by umadevi9616
</script>

Output

Level Order traversal of binary tree is 
1 2 3 4 5 

Time Complexity: O(N2), where N is the number of nodes in the skewed tree. So time complexity of printLevelOrder() is O(n) + O(n-1) + O(n-2) + .. + O(1) which is O(N2). 
Auxiliary Space:  O(N) in the worst case. For a skewed tree, printGivenLevel() uses O(n) space for the call stack. For a Balanced tree, the call stack uses O(log n) space, (i.e., the height of the balanced tree). 

Level Order Binary Tree Traversal Using Queue

For each node, first, the node is visited and then it’s child nodes are put in a FIFO queue. Then again the first node is popped out and the it’s child nodes are put in a FIFO queue and repeat until que becomes empty.

Follow the below steps to Implement the above idea:

  • Create an empty queue q and push root in q.
  • Run While loop until q is not empty. 
    • Initialize temp_node = q.front() and print temp_node->data.
    • Push temp_node’s children i.e. temp_node -> left then temp_node -> right to q
    • Pop front node from q.

Below is the Implementation of the above approach:

C++




/* C++ program to print level
    order traversal using STL */
#include <bits/stdc++.h>
using namespace std;
 
// A Binary Tree Node
struct Node {
    int data;
    struct Node *left, *right;
};
 
// Iterative method to find height of Binary Tree
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) {
        // Print front of queue and remove it from queue
        Node* node = q.front();
        cout << node->data << " ";
        q.pop();
 
        /* Enqueue left child */
        if (node->left != NULL)
            q.push(node->left);
 
        /*Enqueue right child */
        if (node->right != NULL)
            q.push(node->right);
    }
}
 
// Utility function to create a new tree node
Node* newNode(int data)
{
    Node* temp = new Node;
    temp->data = data;
    temp->left = temp->right = NULL;
    return temp;
}
 
// Driver program to test above functions
int main()
{
    // Let us create binary tree shown in above diagram
    Node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);
 
    cout << "Level Order traversal of binary tree is \n";
    printLevelOrder(root);
    return 0;
}

C




// Iterative Queue based C program
// to do level order traversal
// of Binary Tree
#include <stdio.h>
#include <stdlib.h>
#define MAX_Q_SIZE 500
 
/* A binary tree node has data,
   pointer to left child
   and a pointer to right child */
struct node {
    int data;
    struct node* left;
    struct node* right;
};
 
/* frunction prototypes */
struct node** createQueue(int*, int*);
void enQueue(struct node**, int*, struct node*);
struct node* deQueue(struct node**, int*);
 
/* Given a binary tree, print its nodes in level order
   using array for implementing queue */
void printLevelOrder(struct node* root)
{
    int rear, front;
    struct node** queue = createQueue(&front, &rear);
    struct node* temp_node = root;
 
    while (temp_node) {
        printf("%d ", temp_node->data);
 
        /*Enqueue left child */
        if (temp_node->left)
            enQueue(queue, &rear, temp_node->left);
 
        /*Enqueue right child */
        if (temp_node->right)
            enQueue(queue, &rear, temp_node->right);
 
        /*Dequeue node and make it temp_node*/
        temp_node = deQueue(queue, &front);
    }
}
 
/*UTILITY FUNCTIONS*/
struct node** createQueue(int* front, int* rear)
{
    struct node** queue = (struct node**)malloc(
        sizeof(struct node*) * MAX_Q_SIZE);
 
    *front = *rear = 0;
    return queue;
}
 
void enQueue(struct node** queue, int* rear,
             struct node* new_node)
{
    queue[*rear] = new_node;
    (*rear)++;
}
 
struct node* deQueue(struct node** queue, int* front)
{
    (*front)++;
    return queue[*front - 1];
}
 
/* Helper function that allocates a new node with the
   given data and NULL left and right pointers. */
struct node* newNode(int data)
{
    struct node* node
        = (struct node*)malloc(sizeof(struct node));
    node->data = data;
    node->left = NULL;
    node->right = NULL;
 
    return (node);
}
 
/* Driver program to test above functions*/
int main()
{
    struct node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);
 
    printf("Level Order traversal of binary tree is \n");
    printLevelOrder(root);
 
    return 0;
}

Java




// Iterative Queue based Java program
// to do level order traversal
// of Binary Tree
 
/* importing the inbuilt java classes
   required for the program */
import java.util.LinkedList;
import java.util.Queue;
 
/* Class to represent Tree node */
class Node {
    int data;
    Node left, right;
 
    public Node(int item)
    {
        data = item;
        left = null;
        right = null;
    }
}
 
/* Class to print Level Order Traversal */
class BinaryTree {
 
    Node root;
 
    /* Given a binary tree. Print
     its nodes in level order
     using array for implementing queue  */
    void printLevelOrder()
    {
        Queue<Node> queue = new LinkedList<Node>();
        queue.add(root);
        while (!queue.isEmpty()) {
 
            /* poll() removes the present head.
            For more information on poll() visit
            http://www.tutorialspoint.com/java/
            util/linkedlist_poll.htm */
            Node tempNode = queue.poll();
            System.out.print(tempNode.data + " ");
 
            /*Enqueue left child */
            if (tempNode.left != null) {
                queue.add(tempNode.left);
            }
 
            /*Enqueue right child */
            if (tempNode.right != null) {
                queue.add(tempNode.right);
            }
        }
    }
 
    public static void main(String args[])
    {
        /* creating a binary tree and entering
         the nodes */
        BinaryTree tree_level = new BinaryTree();
        tree_level.root = new Node(1);
        tree_level.root.left = new Node(2);
        tree_level.root.right = new Node(3);
        tree_level.root.left.left = new Node(4);
        tree_level.root.left.right = new Node(5);
 
        System.out.println("Level order traversal of binary tree is - ");
        tree_level.printLevelOrder();
    }
}

Python3




# Python program to print level
# order traversal using Queue
 
# A node structure
 
 
class Node:
    # A utility function to create a new node
    def __init__(self, key):
        self.data = key
        self.left = None
        self.right = None
 
# Iterative Method to print the
# height of a binary tree
 
 
def printLevelOrder(root):
    # Base Case
    if root is None:
        return
 
    # Create an empty queue
    # for level order traversal
    queue = []
 
    # Enqueue Root and initialize height
    queue.append(root)
 
    while(len(queue) > 0):
 
        # Print front of queue and
        # remove it from queue
        print(queue[0].data, end = " ")
        node = queue.pop(0)
 
        # Enqueue left child
        if node.left is not None:
            queue.append(node.left)
 
        # Enqueue right child
        if node.right is not None:
            queue.append(node.right)
 
 
# Driver Program to test above function
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.left.right = Node(5)
 
print("Level Order Traversal of binary tree is -")
printLevelOrder(root)
# This code is contributed by Nikhil Kumar Singh(nickzuck_007)

C#




// Iterative Queue based C# program
// to do level order traversal
// of Binary Tree
 
using System;
using System.Collections.Generic;
 
/* Class to represent Tree node */
public class Node {
    public int data;
    public Node left, right;
 
    public Node(int item)
    {
        data = item;
        left = null;
        right = null;
    }
}
 
/* Class to print Level Order Traversal */
public class BinaryTree {
 
    Node root;
 
    /* Given a binary tree. Print
    its nodes in level order using
     array for implementing queue */
    void printLevelOrder()
    {
        Queue<Node> queue = new Queue<Node>();
        queue.Enqueue(root);
        while (queue.Count != 0) {
 
            Node tempNode = queue.Dequeue();
            Console.Write(tempNode.data + " ");
 
            /*Enqueue left child */
            if (tempNode.left != null) {
                queue.Enqueue(tempNode.left);
            }
 
            /*Enqueue right child */
            if (tempNode.right != null) {
                queue.Enqueue(tempNode.right);
            }
        }
    }
 
    // Driver code
    public static void Main()
    {
        /* creating a binary tree and entering
        the nodes */
        BinaryTree tree_level = new BinaryTree();
        tree_level.root = new Node(1);
        tree_level.root.left = new Node(2);
        tree_level.root.right = new Node(3);
        tree_level.root.left.left = new Node(4);
        tree_level.root.left.right = new Node(5);
 
        Console.WriteLine("Level order traversal "
                          + "of binary tree is - ");
        tree_level.printLevelOrder();
    }
}
 
/* This code contributed by PrinciRaj1992 */

Javascript




<script>
// Iterative Queue based javascript program
// to do level order traversal
// of Binary Tree
 
/* Class to represent Tree node */
class Node {
    constructor(val) {
        this.data = val;
        this.left = null;
        this.right = null;
    }
}
 
/* Class to print Level Order Traversal */
    /*
     * Given a binary tree. Print its nodes in level order using array for
     * implementing queue
     */
    function printLevelOrder() {
        var queue = [];
        queue.push(root);
        while (queue.length != 0) {
            /*
             * The shift() method removes the first element from an array and returns that removed element. This method changes the length of the array.
             */
            var tempNode = queue.shift();
            document.write(tempNode.data + " ");
 
            /* Enqueue left child */
            if (tempNode.left != null) {
                queue.push(tempNode.left);
            }
 
            /* Enqueue right child */
            if (tempNode.right != null) {
                queue.push(tempNode.right);
            }
        }
    }
 
  /* creating a binary tree and entering the nodes */
        var 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);
        document.write("Level order traversal of binary tree is - ");
        printLevelOrder();
 
// This code is contributed by umadevi9616
</script>

Output

Level Order traversal of binary tree is 
1 2 3 4 5 

Time Complexity: O(N) where n is the number of nodes in the binary tree.
Auxiliary Space: O(N) where n is the number of nodes in the binary tree.

Please write comments if you find any bugs in the above programs/algorithms or other ways to solve the same problem. 


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