Get Level of a node in a Binary Tree
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 ispresent 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 anew Binary Tree node */struct 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 " << getLevel(root, x) << endl; else cout << x << "is not present in tree" << endl; } getchar(); return 0;}// This code is contributed// by Akanksha Rai |
C
// C program to Get Level of a node in a Binary Tree#include<stdio.h>#include<stdlib.h>/* 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 node */struct node* newNode(int data){ struct node *temp = (struct node*)malloc(sizeof(struct node)); temp->data = data; temp->left = NULL; temp->right = NULL; return temp;}/* Driver code */int main(){ struct 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 has been contributed by Mayank// Jaiswal(mayank_24) |
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 |
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 of getLevel() is O(n) where n is the number of nodes in the given Binary Tree.
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