Level of a Node in Binary Tree
Last Updated :
29 Sep, 2024
Improve
Try it on GfG Practice 
Given a Binary Tree and a key, the task is to find the level of key in the Binary Tree.
Examples:
Input : key = 4
Output: 3
Explanation: The level of the key in above binary tree is 3.
Input : key = 10Output: -1
Explanation: Key is not present in the above Binary tree.
Table of Content
[Expected Approach - 1] Using Recursion - O(n) Time and O(h) Space
The idea is to start from the root and level as 1. If the target matches with root's data, return level. Else recursively call for left and right subtrees with level as level + 1.
Below is the implementation of the above approach:
// C++ code to find level of a Node in Binary Tree
#include <iostream>
using namespace std;
class Node {
public:
int data;
Node* left;
Node* right;
Node(int val) {
data = val;
left = right = nullptr;
}
};
// Recursive function to find the level of the target key
int getLevel(Node* root, int target, int level) {
if (root == nullptr) {
return -1;
}
// If the target key matches the current node's
// data, return the level
if (root->data == target) {
return level;
}
// Recursively call for left and right subtrees
int leftLevel = getLevel(root->left, target, level + 1);
if (leftLevel != -1) {
return leftLevel;
}
return getLevel(root->right, target, level + 1);
}
int main() {
// Creating a sample binary tree:
// 1
// / \
// 2 3
// / \ / \
// 4 5 6 7
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->left = new Node(6);
root->right->right = new Node(7);
int target = 5;
cout << getLevel(root, target, 1) << endl;
return 0;
}
// C code to find level of a Node in Binary Tree
#include <stdio.h>
#include <stdlib.h>
struct Node {
int data;
struct Node* left;
struct Node* right;
};
int getLevel(struct Node* root, int target, int level) {
if (root == NULL) {
return -1;
}
// If the target key matches the current node's data, return the level
if (root->data == target) {
return level;
}
// Recursively call for left and right subtrees
int leftLevel = getLevel(root->left, target, level + 1);
if (leftLevel != -1) {
return leftLevel;
}
return getLevel(root->right, target, level + 1);
}
struct Node* createNode(int val) {
struct Node* newNode =
(struct Node*)malloc(sizeof(struct Node));
newNode->data = val;
newNode->left = newNode->right = NULL;
return newNode;
}
int main() {
// Creating a sample binary tree:
// 1
// / \
// 2 3
// / \ / \
// 4 5 6 7
struct Node* root = createNode(1);
root->left = createNode(2);
root->right = createNode(3);
root->left->left = createNode(4);
root->left->right = createNode(5);
root->right->left = createNode(6);
root->right->right = createNode(7);
int target = 5;
printf("%d\n", getLevel(root, target, 1));
return 0;
}
// Java code to find level of a Node in Binary Tree
class Node {
int data;
Node left, right;
Node(int val) {
data = val;
left = right = null;
}
}
class GfG {
// Recursive function to find the level of the target key
static int getLevel(Node root, int target, int level) {
if (root == null) {
return -1;
}
// If the target key matches the current node's
// data, return the level
if (root.data == target) {
return level;
}
// Recursively call for left and right subtrees
int leftLevel = getLevel(root.left, target, level + 1);
if (leftLevel != -1) {
return leftLevel;
}
return getLevel(root.right, target, level + 1);
}
public static void main(String[] args) {
// Creating a sample binary tree:
// 1
// / \
// 2 3
// / \ / \
// 4 5 6 7
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.left = new Node(6);
root.right.right = new Node(7);
int target = 5;
System.out.println(getLevel(root, target, 1));
}
}
# Python code to find level of a Node in Binary Tree
class Node:
def __init__(self, val):
self.data = val
self.left = None
self.right = None
# Recursive function to find the level of the target key
def getLevel(root, target, level):
if root is None:
return -1
# If the target key matches the current node's
# data, return the level
if root.data == target:
return level
# Recursively call for left and right subtrees
leftLevel = getLevel(root.left, target, level + 1)
if leftLevel != -1:
return leftLevel
return getLevel(root.right, target, level + 1)
if __name__ == "__main__":
# Creating a sample binary tree:
# 1
# / \
# 2 3
# / \ / \
# 4 5 6 7
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.left.right = Node(5)
root.right.left = Node(6)
root.right.right = Node(7)
target = 5
print(getLevel(root, target, 1))
// C# code to find level of a Node in Binary Tree
using System;
class Node {
public int Data;
public Node Left;
public Node Right;
public Node(int val) {
Data = val;
Left = Right = null;
}
}
class GfG {
// Recursive function to find the level of the target key
static int GetLevel(Node root, int target, int level) {
if (root == null) {
return -1;
}
// If the target key matches the current
// node's data, return the level
if (root.Data == target) {
return level;
}
// Recursively call for left and right subtrees
int leftLevel = GetLevel(root.Left, target, level + 1);
if (leftLevel != -1) {
return leftLevel;
}
return GetLevel(root.Right, target, level + 1);
}
static void Main() {
// Creating a sample binary tree:
// 1
// / \
// 2 3
// / \ / \
// 4 5 6 7
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.Left = new Node(6);
root.Right.Right = new Node(7);
int target = 5;
Console.WriteLine(GetLevel(root, target, 1));
}
}
// Javascript code to find level of a Node
// in Binary Tree
class Node {
constructor(val) {
this.data = val;
this.left = null;
this.right = null;
}
}
// Recursive function to find the level of the target key
function getLevel(root, target, level) {
if (root === null) {
return -1;
}
// If the target key matches the current node's
// data, return the level
if (root.data === target) {
return level;
}
// Recursively call for left and right subtrees
let leftLevel = getLevel(root.left, target, level + 1);
if (leftLevel !== -1) {
return leftLevel;
}
return getLevel(root.right, target, level + 1);
}
// Creating a sample binary tree:
// 1
// / \
// 2 3
// / \ / \
// 4 5 6 7
const 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.left = new Node(6);
root.right.right = new Node(7);
const target = 5;
console.log(getLevel(root, target, 1));
Output
3
Time Complexity: O(n), where n is the number of nodes in the binary tree.
Auxiliary Space: O(h), where h is height of binary tree.
[Expected Approach - 2] Using Level Order Traversal- O(n) Time and O(n) Space
The idea is to perform a level-order traversal and keep track of the current level as we traverse the tree. If the key matches with root's data, return level.
// C++ code to find level of a Node in Binary Tree
#include <bits/stdc++.h>
using namespace std;
class Node {
public:
int data;
Node* left;
Node* right;
Node(int val) {
data = val;
left = right = nullptr;
}
};
// Function to find the level of the target key
int getLevel(Node* root, int target) {
if (root == nullptr) {
return -1;
}
// Create a queue for level-order
// traversal
queue<Node*> q;
q.push(root);
int level = 1;
while (!q.empty()) {
int size = q.size();
// Process all nodes at the current level
for (int i = 0; i < size; i++) {
Node* curr = q.front();
q.pop();
// Check if the current node matches the target
if (curr->data == target) {
return level;
}
// Push the left and right children to the queue
if (curr->left != nullptr) {
q.push(curr->left);
}
if (curr->right != nullptr) {
q.push(curr->right);
}
}
level++;
}
return -1;
}
int main() {
// Creating a sample binary tree:
// 1
// / \
// 2 3
// / \ / \
// 4 5 6 7
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->left = new Node(6);
root->right->right = new Node(7);
int target = 5;
cout << getLevel(root, target);
return 0;
}
// Java code to find level of a Node
// in Binary Tree
import java.util.LinkedList;
import java.util.Queue;
class Node {
int data;
Node left, right;
Node(int val) {
data = val;
left = right = null;
}
}
// Function to find the level of
// the target key
class GfG {
static int getLevel(Node root, int target) {
if (root == null) {
return -1;
}
// Create a queue for level-order traversal
Queue<Node> q = new LinkedList<>();
q.add(root);
int level = 1;
while (!q.isEmpty()) {
int size = q.size();
// Process all nodes at the current level
for (int i = 0; i < size; i++) {
Node curr = q.poll();
// Check if the current node matches the target
if (curr.data == target) {
return level;
}
// Push the left and right children to the queue
if (curr.left != null) {
q.add(curr.left);
}
if (curr.right != null) {
q.add(curr.right);
}
}
level++;
}
return -1;
}
public static void main(String[] args) {
// Creating a sample binary tree:
// 1
// / \
// 2 3
// / \ / \
// 4 5 6 7
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.left = new Node(6);
root.right.right = new Node(7);
int target = 5;
System.out.println(getLevel(root, target));
}
}
# Python code to find level of a Node in Binary Tree
class Node:
def __init__(self, val):
self.data = val
self.left = None
self.right = None
# Function to find the level of the target key
def getLevel(root, target):
if root is None:
return -1
# Create a queue for level-order traversal
q = []
q.append(root)
level = 1
while len(q) > 0:
size = len(q)
# Process all nodes at the current level
for i in range(size):
curr = q.pop(0)
# Check if the current node matches the target
if curr.data == target:
return level
# Push the left and right children to the queue
if curr.left is not None:
q.append(curr.left)
if curr.right is not None:
q.append(curr.right)
level += 1
return -1
if __name__ == "__main__":
# Creating a sample binary tree:
# 1
# / \
# 2 3
# / \ / \
# 4 5 6 7
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.left.right = Node(5)
root.right.left = Node(6)
root.right.right = Node(7)
target = 5
print(getLevel(root, target))
// C# code to find level of a Node in Binary Tree
using System;
using System.Collections.Generic;
class Node {
public int data;
public Node left, right;
public Node(int val) {
data = val;
left = right = null;
}
}
class GfG {
// Function to find the level of the target key
static int GetLevel(Node root, int target) {
if (root == null) {
return -1;
}
// Create a queue for level-order traversal
Queue<Node> q = new Queue<Node>();
q.Enqueue(root);
int level = 1;
while (q.Count > 0) {
int size = q.Count;
// Process all nodes at the current level
for (int i = 0; i < size; i++) {
Node curr = q.Dequeue();
// Check if the current node matches the target
if (curr.data == target) {
return level;
}
// Push the left and right children to the queue
if (curr.left != null) {
q.Enqueue(curr.left);
}
if (curr.right != null) {
q.Enqueue(curr.right);
}
}
level++;
}
return -1;
}
static void Main(string[] args) {
// Creating a sample binary tree:
// 1
// / \
// 2 3
// / \ / \
// 4 5 6 7
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.left = new Node(6);
root.right.right = new Node(7);
int target = 5;
Console.WriteLine(GetLevel(root, target));
}
}
// Javascript code to find level of
// a Node in Binary Tree
class Node {
constructor(val) {
this.data = val;
this.left = this.right = null;
}
}
// Function to find the level of the target key
function getLevel(root, target) {
if (root === null) {
return -1;
}
// Create a queue for level-order traversal
let q = [];
q.push(root);
let level = 1;
while (q.length > 0) {
let size = q.length;
// Process all nodes at the current level
for (let i = 0; i < size; i++) {
let curr = q.shift();
// Check if the current node matches the target
if (curr.data === target) {
return level;
}
// Push the left and right children to the queue
if (curr.left !== null) {
q.push(curr.left);
}
if (curr.right !== null) {
q.push(curr.right);
}
}
level++;
}
return -1;
}
// Creating the binary tree
// 1
// / \
// 2 3
// / \ / \
// 4 5 6 7
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.left = new Node(6);
root.right.right = new Node(7);
let target = 5;
console.log(getLevel(root, target));
Output
3
Time Complexity: O(n), where n is the number of nodes in the binary tree.
Auxiliary Space: O(n)
