Level Order Binary Tree Traversal

• Difficulty Level : Easy
• Last Updated : 30 Dec, 2021

Level order traversal of a tree is breadth first traversal for the tree. Level order traversal of the above tree is 1 2 3 4 5

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Method 1 (Use function to print a current level)

Algorithm:
There are basically two functions in this method. One is to print all nodes at a given level (printCurrentLevel), and other is to print level order traversal of the tree (printLevelorder). printLevelorder makes use of printCurrentLevel to print nodes at all levels one by one starting from the root.

/*Function to print level order traversal of tree*/
printLevelorder(tree)
for d = 1 to height(tree)
printCurrentLevel(tree, d);

/*Function to print all nodes at a current level*/
printCurrentLevel(tree, level)
if tree is NULL then return;
if level is 1, then
print(tree->data);
else if level greater than 1, then
printCurrentLevel(tree->left, level-1);
printCurrentLevel(tree->right, level-1);

Implementation:

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(n^2) in worst case. For a skewed tree, printGivenLevel() takes O(n) time 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(n^2).
Space Complexity: O(n) in worst case. For a skewed tree, printGivenLevel() uses O(n) space for call stack. For a Balanced tree, the call stack uses O(log n) space, (i.e., the height of the balanced tree).

Method 2 (Using queue)

Algorithm:
For each node, first the node is visited and then it’s child nodes are put in a FIFO queue.

printLevelorder(tree)
1) Create an empty queue q
2) temp_node = root /*start from root*/
3) Loop while temp_node is not NULL
a) print temp_node->data.
b) Enqueue temp_node’s children
(first left then right children) to q
c) Dequeue a node from q.

Implementation:
Here is a simple implementation of the above algorithm. Queue is implemented using an array with a maximum size of 500. We can implement queue as a linked list also.

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.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()
{
while (!queue.isEmpty()) {

/* poll() removes the present head.
http://www.tutorialspoint.com/java/
Node tempNode = queue.poll();
System.out.print(tempNode.data + " ");

/*Enqueue left child */
if (tempNode.left != null) {
}

/*Enqueue right child */
if (tempNode.right != null) {
}
}
}

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.data)
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

/* importing the inbuilt javascript classes
required for the program */

/* 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) {

/*
*/
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
Space Complexity: O(n) where n is the number of nodes in the binary tree

References: