Given a Binary Tree and a key, write a function that returns level of the key.
For example, consider the following tree. If the input key is 3, then your function should return 1. If the input key is 4, then your function should return 3. And for key which is not present in key, then your function should return 0.
The idea is to start from the root and level as 1. If the key matches with root’s data, return level. Else recursively call for left and right subtrees with level as level + 1.
C++
// C++ program to Get Level of a node in a Binary Tree#include <bits/stdc++.h>using namespace std;/* A tree node structure */struct node { int data; struct node* left; struct node* right;};// Helper function for getLevel(). It returns level of the// data if data is present in tree, otherwise returns 0.int getLevelUtil(struct node* node, int data, int level){ if (node == NULL) return 0; if (node->data == data) return level; int downlevel = getLevelUtil(node->left, data, level + 1); if (downlevel != 0) return downlevel; downlevel = getLevelUtil(node->right, data, level + 1); return downlevel;}/* Returns level of given data value */int getLevel(struct node* node, int data){ return getLevelUtil(node, data, 1);}// Utility function to create a new Binary Tree nodestruct node* newNode(int data){ struct node* temp = new struct node; temp->data = data; temp->left = NULL; temp->right = NULL; return temp;}// Driver Codeint main(){ struct node* root = new struct node; int x; // Constructing tree given in the above figure root = newNode(3); root->left = newNode(2); root->right = newNode(5); root->left->left = newNode(1); root->left->right = newNode(4); for (x = 1; x <= 5; x++) { int level = getLevel(root, x); if (level) cout << "Level of " << x << " is " << level << endl; else cout << x << "is not present in tree" << endl; } getchar(); return 0;}// This code is contributed by Aditya Kumar (adityakumar129) |
C
// C program to Get Level of a node in a Binary Tree#include <stdio.h>#include <stdlib.h>/* A tree node structure */typedef struct node { int data; struct node* left; struct node* right;} node;// Helper function for getLevel(). It returns level of the// data if data is present in tree, otherwise returns 0.int getLevelUtil(node* node, int data, int level){ if (node == NULL) return 0; if (node->data == data) return level; int downlevel = getLevelUtil(node->left, data, level + 1); if (downlevel != 0) return downlevel; downlevel = getLevelUtil(node->right, data, level + 1); return downlevel;}/* Returns level of given data value */int getLevel(node* node, int data){ return getLevelUtil(node, data, 1);}// Utility function to create a new Binary Tree nodenode* newNode(int data){ node* temp = (node*)malloc(sizeof(node)); temp->data = data; temp->left = NULL; temp->right = NULL; return temp;}/* Driver code */int main(){ node* root; int x; // Constructing tree given in the above figure root = newNode(3); root->left = newNode(2); root->right = newNode(5); root->left->left = newNode(1); root->left->right = newNode(4); for (x = 1; x <= 5; x++) { int level = getLevel(root, x); if (level) printf(" Level of %d is %d\n", x, getLevel(root, x)); else printf(" %d is not present in tree \n", x); } getchar(); return 0;} |
Java
// Java program to Get Level of a node in a Binary Tree/* A tree node structure */class Node { int data; Node left, right; public Node(int d) { data = d; left = right = null; }}class BinaryTree { Node root; // Helper function for getLevel(). It returns level of // the data if data is present in tree, otherwise // returns 0. int getLevelUtil(Node node, int data, int level) { if (node == null) return 0; if (node.data == data) return level; int downlevel = getLevelUtil(node.left, data, level + 1); if (downlevel != 0) return downlevel; downlevel = getLevelUtil(node.right, data, level + 1); return downlevel; } /* Returns level of given data value */ int getLevel(Node node, int data) { return getLevelUtil(node, data, 1); } /* Driver code */ public static void main(String[] args) { BinaryTree tree = new BinaryTree(); /* Constructing tree given in the above figure */ tree.root = new Node(3); tree.root.left = new Node(2); tree.root.right = new Node(5); tree.root.left.left = new Node(1); tree.root.left.right = new Node(4); for (int x = 1; x <= 5; x++) { int level = tree.getLevel(tree.root, x); if (level != 0) System.out.println( "Level of " + x + " is " + tree.getLevel(tree.root, x)); else System.out.println( x + " is not present in tree"); } }}// This code is contributed by Aditya Kumar (adityakumar129) |
Python3
# Python3 program to Get Level of a# node in a Binary Tree# Helper function that allocates a# new node with the given data and# None left and right pairs.class newNode: # Constructor to create a new node def __init__(self, data): self.data = data self.left = None self.right = None# Helper function for getLevel(). It# returns level of the data if data is# present in tree, otherwise returns 0def getLevelUtil(node, data, level): if (node == None): return 0 if (node.data == data): return level downlevel = getLevelUtil(node.left, data, level + 1) if (downlevel != 0): return downlevel downlevel = getLevelUtil(node.right, data, level + 1) return downlevel# Returns level of given data valuedef getLevel(node, data): return getLevelUtil(node, data, 1)# Driver Codeif __name__ == '__main__': # Let us construct the Tree shown # in the above figure root = newNode(3) root.left = newNode(2) root.right = newNode(5) root.left.left = newNode(1) root.left.right = newNode(4) for x in range(1, 6): level = getLevel(root, x) if (level): print("Level of", x, "is", getLevel(root, x)) else: print(x, "is not present in tree")# This code is contributed by# Shubham Singh(SHUBHAMSINGH10) |
C#
// C# program to Get Level of a node in a Binary Treeusing System;/* A tree node structure */public class Node { public int data; public Node left, right; public Node(int d) { data = d; left = right = null; }}public class BinaryTree { public Node root; /* Helper function for getLevel(). It returns level of the data if data is present in tree, otherwise returns 0.*/ public virtual int getLevelUtil(Node node, int data, int level) { if (node == null) { return 0; } if (node.data == data) { return level; } int downlevel = getLevelUtil(node.left, data, level + 1); if (downlevel != 0) { return downlevel; } downlevel = getLevelUtil(node.right, data, level + 1); return downlevel; } /* Returns level of given data value */ public virtual int getLevel(Node node, int data) { return getLevelUtil(node, data, 1); } /* Driver code */ public static void Main(string[] args) { BinaryTree tree = new BinaryTree(); /* Constructing tree given in the above figure */ tree.root = new Node(3); tree.root.left = new Node(2); tree.root.right = new Node(5); tree.root.left.left = new Node(1); tree.root.left.right = new Node(4); for (int x = 1; x <= 5; x++) { int level = tree.getLevel(tree.root, x); if (level != 0) { Console.WriteLine( "Level of " + x + " is " + tree.getLevel(tree.root, x)); } else { Console.WriteLine( x + " is not present in tree"); } } }}// This code is contributed by Shrikant13 |
Javascript
<script> // Javascript program to Get Level of // a node in a Binary Tree /* A tree node structure */ class Node { constructor(d) { this.data = d; this.left = null; this.right = null; } } let root; /* Helper function for getLevel(). It returns level of the data if data is present in tree, otherwise returns 0.*/ function getLevelUtil(node, data, level) { if (node == null) return 0; if (node.data == data) return level; let downlevel = getLevelUtil(node.left, data, level + 1); if (downlevel != 0) return downlevel; downlevel = getLevelUtil(node.right, data, level + 1); return downlevel; } /* Returns level of given data value */ function getLevel(node, data) { return getLevelUtil(node, data, 1); } /* Constructing tree given in the above figure */ root = new Node(3); root.left = new Node(2); root.right = new Node(5); root.left.left = new Node(1); root.left.right = new Node(4); for (let x = 1; x <= 5; x++) { let level = getLevel(root, x); if (level != 0) document.write( "Level of " + x + " is " + getLevel(root, x) + "</br>"); else document.write( x + " is not present in tree"); } </script> |
Output
Level of 1 is 3 Level of 2 is 2 Level of 3 is 1 Level of 4 is 3 Level of 5 is 2
Time Complexity: O(n) where n is the number of nodes in the given Binary Tree.
Auxiliary Space: O(n)
Alternative Approach: The given problem can be solved with the help of level order traversal of given binary tree.
C++
// C++ program to print level in which X is present in// binary tree using STL#include <bits/stdc++.h>using namespace std;// A Binary Tree Nodestruct Node { int data; struct Node *left, *right;};int printLevel(Node* root, int X){ Node* node; // Base Case if (root == NULL) return 0; // Create an empty queue for level order traversal queue<Node*> q; // Create a var represent current level of tree int currLevel = 1; // Enqueue Root q.push(root); while (q.empty() == false) { int size = q.size(); while (size--) { node = q.front(); if (node->data == X) return currLevel; q.pop(); /* Enqueue left child */ if (node->left != NULL) q.push(node->left); /*Enqueue right child */ if (node->right != NULL) q.push(node->right); } currLevel++; } return 0;}// Utility function to create a new tree nodeNode* newNode(int data){ Node* temp = new Node; temp->data = data; temp->left = temp->right = NULL; return temp;}// Driver program to test above functionsint 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); root->right->left = newNode(7); root->right->right = newNode(6); cout << printLevel(root, 6); return 0;}// This code is contributed by Aditya Kumar (adityakumar129) |
Java
// Java program to print level in which X is present in// binary treeimport 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. // Create a var represent current level of tree public static int currLevel = 1; int printLevelOrder(int X) { // Create an empty queue for level order traversal Queue<Node> queue = new LinkedList<Node>(); queue.add(root); while (!queue.isEmpty()) { // poll() removes the present head. For more // information on poll() visit int size = queue.size(); for (int i = 0; i < size; i++) { Node tempNode = queue.poll(); if (tempNode.data == X) return currLevel; /*Enqueue left child */ if (tempNode.left != null) queue.add(tempNode.left); /*Enqueue right child */ if (tempNode.right != null) queue.add(tempNode.right); } currLevel++; } return currLevel; } 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); tree_level.root.right.left = new Node(7); tree_level.root.right.right = new Node(6); System.out.println( "Level order traversal of binary tree is - " + tree_level.printLevelOrder(6)); }}// This code is contributed by Aditya Kumar (adityakumar129) |
Python3
# Python3 program to print level in which X is present in# binary tree# A node structureclass Node: # A utility function to create a new node def __init__(self, key): self.data = key self.left = None self.right = Nonedef printLevel(root, X): # Base Case if root is None: return 0 # Create an empty queue # for level order traversal q = [] #Create a var represent current level of tree currLevel = 1 # Enqueue Root q.append(root) while(len(q) > 0): size = len(q) for i in range(size): node = q.pop(0) if(node.data == X): return currLevel # Enqueue left child if node.left is not None: q.append(node.left) # Enqueue right child if node.right is not None: q.append(node.right) currLevel += 1 return 0# Driver Program to test above functionroot = Node(1)root.left = Node(2)root.right = Node(3)root.left.left = Node(4)root.left.right = Node(5)root.right.left = Node(7)root.right.right = Node(6)print(printLevel(root, 6))# This code is contributed by Abhijeet Kumar(abhijeet19403) |
C#
// C# program to print level in which X is present in// binary treeusing 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; /* Function to get the level of the node X*/ int printLevel(int X) { Node node; // Base Case if (root == null) return 0; // Create an empty queue for level order traversal Queue<Node> q = new Queue<Node>(); // Create a var represent current level of tree int currLevel = 1; // Enqueue root q.Enqueue(root); while (q.Count != 0) { int size = q.Count; while (size-- != 0) { node = q.Dequeue(); if (node.data == X) return currLevel; /*Enqueue left child */ if (node.left != null) q.Enqueue(node.left); /*Enqueue right child */ if (node.right != null) q.Enqueue(node.right); } currLevel++; } return 0; } // 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); tree_level.root.right.left = new Node(7); tree_level.root.right.right = new Node(6); Console.WriteLine(tree_level.printLevel(6)); }}/* This code contributed by Abhijeet Kumar(abhijeet19403) */ |
Javascript
// JavaScript program to print level in which X is present in// binary treeclass Node{ constructor(d) { this.data = d; this.left = null; this.right = null; }}let root;// Given a binary tree. Print its nodes in level order// using array for implementing queue.// Create a var represent current level of treelet currLevel = 1;function printLevelOrder(X){ // Create an empty queue for level order traversal let queue = [root]; while(queue[0]){ let size = queue.length; for(let i=0;i<size;i++){ let tempNode = queue.shift(); if(tempNode.data==X){ return currLevel; } /*Enqueue left child */ if(tempNode.left){ queue.push(tempNode.left); } /*Enqueue right child */ if(tempNode.right){ queue.push(tempNode.right); } } currLevel+=1; } return root.data;}/* creating a binary tree and entering the nodes */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(7);root.right.right = new Node(6);console.log(printLevelOrder(6));// This code is contributed by lokeshmvs21. |
Output
3
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 anything incorrect, or you want to share more information about the topic discussed above.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
