Given a Binary Tree consisting of N nodes and a integer K, the task is to find the depth and height of the node with value K in the Binary Tree.
The depth of a node is the number of edges present in path from the root node of a tree to that node.
The height of a node is the number of edges present in the longest path connecting that node to a leaf node.
Examples:
Input: K = 25,
5
/ \
10 15
/ \ / \
20 25 30 35
\
45
Output:
Depth of node 25 = 2
Height of node 25 = 1
Explanation:
The number of edges in the path from root node to the node 25 is 2. Therefore, depth of the node 25 is 2.
The number of edges in the longest path connecting the node 25 to any leaf node is 1. Therefore, height of the node 25 is 1.Input: K = 10,
5
/ \
10 15
/ \ / \
20 25 30 35
\
45
Output:
Depth of node 10 = 1
Height of node 10 = 2
Approach: The problem can be solved based on the following observations:
Depth of a node K (of a Binary Tree) = Number of edges in the path connecting the root to the node K = Number of ancestors of K (excluding K itself).
Follow the steps below to find the depth of the given node:
- If the tree is empty, print -1.
- Otherwise, perform the following steps:
- Initialize a variable, say dist as -1.
- Check if the node K is equal to the given node.
- Otherwise, check if it is present in either of the subtrees, by recursively checking for the left and right subtrees respectively.
- If found to be true, print the value of dist + 1.
- Otherwise, print dist.
Height of a node K (of a Binary Tree) = Number of edges in the longest path connecting K to any leaf node.
Follow the steps below to find the height of the given node:
- If the tree is empty, print -1.
- Otherwise, perform the following steps:
- Calculate the height of the left subtree recursively.
- Calculate the height of the right subtree recursively.
- Update height of the current node by adding 1 to the maximum of the two heights obtained in the previous step. Store the height in a variable, say ans.
- If the current node is equal to the given node K, print the value of ans as the required answer.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Structure of a Binary Tree Nodestruct Node { int data; Node *left, *right;};// Utility function to create// a new Binary Tree NodeNode* newNode(int item){ Node* temp = new Node; temp->data = item; temp->left = temp->right = NULL; return temp;}// Function to find the depth of// a given node in a Binary Treeint findDepth(Node* root, int x){ // Base case if (root == NULL) return -1; // Initialize distance as -1 int dist = -1; // Check if x is current node= if ((root->data == x) // Otherwise, check if x is // present in the left subtree || (dist = findDepth(root->left, x)) >= 0 // Otherwise, check if x is // present in the right subtree || (dist = findDepth(root->right, x)) >= 0) // Return depth of the node return dist + 1; return dist;}// Helper function to find the height// of a given node in the binary treeint findHeightUtil(Node* root, int x, int& height){ // Base Case if (root == NULL) { return -1; } // Store the maximum height of // the left and right subtree int leftHeight = findHeightUtil( root->left, x, height); int rightHeight = findHeightUtil( root->right, x, height); // Update height of the current node int ans = max(leftHeight, rightHeight) + 1; // If current node is the required node if (root->data == x) height = ans; return ans;}// Function to find the height of// a given node in a Binary Treeint findHeight(Node* root, int x){ // Store the height of // the given node int h = -1; // Stores height of the Tree int maxHeight = findHeightUtil(root, x, h); // Return the height return h;}// Driver Codeint main(){ // Binary Tree Formation Node* root = newNode(5); root->left = newNode(10); root->right = newNode(15); root->left->left = newNode(20); root->left->right = newNode(25); root->left->right->right = newNode(45); root->right->left = newNode(30); root->right->right = newNode(35); int k = 25; // Function call to find the // depth of a given node cout << "Depth: " << findDepth(root, k) << "\n"; // Function call to find the // height of a given node cout << "Height: " << findHeight(root, k); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{static int height = -1;// Structure of a Binary Tree Nodestatic class Node { int data; Node left; Node right;};// Utility function to create// a new Binary Tree Nodestatic Node newNode(int item){ Node temp = new Node(); temp.data = item; temp.left = temp.right = null; return temp;}// Function to find the depth of// a given node in a Binary Treestatic int findDepth(Node root, int x){ // Base case if (root == null) return -1; // Initialize distance as -1 int dist = -1; // Check if x is current node= if ((root.data == x)|| // Otherwise, check if x is // present in the left subtree (dist = findDepth(root.left, x)) >= 0 || // Otherwise, check if x is // present in the right subtree (dist = findDepth(root.right, x)) >= 0) // Return depth of the node return dist + 1; return dist;}// Helper function to find the height// of a given node in the binary treestatic int findHeightUtil(Node root, int x){ // Base Case if (root == null) { return -1; } // Store the maximum height of // the left and right subtree int leftHeight = findHeightUtil(root.left, x); int rightHeight = findHeightUtil(root.right, x); // Update height of the current node int ans = Math.max(leftHeight, rightHeight) + 1; // If current node is the required node if (root.data == x) height = ans; return ans;}// Function to find the height of// a given node in a Binary Treestatic int findHeight(Node root, int x){ // Stores height of the Tree findHeightUtil(root, x); // Return the height return height;}// Driver Codepublic static void main(String []args){ // Binary Tree Formation Node root = newNode(5); root.left = newNode(10); root.right = newNode(15); root.left.left = newNode(20); root.left.right = newNode(25); root.left.right.right = newNode(45); root.right.left = newNode(30); root.right.right = newNode(35); int k = 25; // Function call to find the // depth of a given node System.out.println("Depth: " + findDepth(root, k)); // Function call to find the // height of a given node System.out.println("Height: " + findHeight(root, k));}}// This code is contributed by SURENDRA_GANGWAR |
Python3
# Python3 program for the above approach# Structure of a Binary Tree Nodeclass Node: def __init__(self, x): self.data = x self.left = None self.right = None # Function to find the depth of# a given node in a Binary Treedef findDepth(root, x): # Base case if (root == None): return -1 # Initialize distance as -1 dist = -1 # Check if x is current node= if (root.data == x): return dist + 1 dist = findDepth(root.left, x) if dist >= 0: return dist + 1 dist = findDepth(root.right, x) if dist >= 0: return dist + 1 return dist# Helper function to find the height# of a given node in the binary treedef findHeightUtil(root, x): global height # Base Case if (root == None): return -1 # Store the maximum height of # the left and right subtree leftHeight = findHeightUtil(root.left, x) rightHeight = findHeightUtil(root.right, x) # Update height of the current node ans = max(leftHeight, rightHeight) + 1 # If current node is the required node if (root.data == x): height = ans return ans# Function to find the height of# a given node in a Binary Treedef findHeight(root, x): global height # Stores height of the Tree maxHeight = findHeightUtil(root, x) # Return the height return height# Driver Codeif __name__ == '__main__': # Binary Tree Formation height = -1 root = Node(5) root.left = Node(10) root.right = Node(15) root.left.left = Node(20) root.left.right = Node(25) root.left.right.right = Node(45) root.right.left = Node(30) root.right.right = Node(35) k = 25 # Function call to find the # depth of a given node print("Depth: ",findDepth(root, k)) # Function call to find the # height of a given node print("Height: ",findHeight(root, k)) # This code is contributed by mohit kumar 29. |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG{static int height = -1;// Structure of a Binary Tree Nodeclass Node { public int data; public Node left; public Node right;};// Utility function to create// a new Binary Tree Nodestatic Node newNode(int item){ Node temp = new Node(); temp.data = item; temp.left = temp.right = null; return temp;}// Function to find the depth of// a given node in a Binary Treestatic int findDepth(Node root, int x){ // Base case if (root == null) return -1; // Initialize distance as -1 int dist = -1; // Check if x is current node= if ((root.data == x)|| // Otherwise, check if x is // present in the left subtree (dist = findDepth(root.left, x)) >= 0 || // Otherwise, check if x is // present in the right subtree (dist = findDepth(root.right, x)) >= 0) // Return depth of the node return dist + 1; return dist;}// Helper function to find the height// of a given node in the binary treestatic int findHeightUtil(Node root, int x){ // Base Case if (root == null) { return -1; } // Store the maximum height of // the left and right subtree int leftHeight = findHeightUtil(root.left, x); int rightHeight = findHeightUtil(root.right, x); // Update height of the current node int ans = Math.Max(leftHeight, rightHeight) + 1; // If current node is the required node if (root.data == x) height = ans; return ans;}// Function to find the height of// a given node in a Binary Treestatic int findHeight(Node root, int x){ // Stores height of the Tree findHeightUtil(root, x); // Return the height return height;}// Driver Codepublic static void Main(){ // Binary Tree Formation Node root = newNode(5); root.left = newNode(10); root.right = newNode(15); root.left.left = newNode(20); root.left.right = newNode(25); root.left.right.right = newNode(45); root.right.left = newNode(30); root.right.right = newNode(35); int k = 25; // Function call to find the // depth of a given node Console.WriteLine("Depth: " + findDepth(root, k)); // Function call to find the // height of a given node Console.WriteLine("Height: " + findHeight(root, k));}}// This code is contributed by ipg2016107 |
Javascript
<script>// JavaScript program for the above approachvar height = -1;// Structure of a Binary Tree Nodeclass Node { constructor() { this.data = 0; this.left = null; this.right = null; }};// Utility function to create// a new Binary Tree Nodefunction newNode(item){ var temp = new Node(); temp.data = item; temp.left = temp.right = null; return temp;}// Function to find the depth of// a given node in a Binary Treefunction findDepth(root, x){ // Base case if (root == null) return -1; // Initialize distance as -1 var dist = -1; // Check if x is current node= if ((root.data == x)|| // Otherwise, check if x is // present in the left subtree (dist = findDepth(root.left, x)) >= 0 || // Otherwise, check if x is // present in the right subtree (dist = findDepth(root.right, x)) >= 0) // Return depth of the node return dist + 1; return dist;}// Helper function to find the height// of a given node in the binary treefunction findHeightUtil(root, x){ // Base Case if (root == null) { return -1; } // Store the maximum height of // the left and right subtree var leftHeight = findHeightUtil(root.left, x); var rightHeight = findHeightUtil(root.right, x); // Update height of the current node var ans = Math.max(leftHeight, rightHeight) + 1; // If current node is the required node if (root.data == x) height = ans; return ans;}// Function to find the height of// a given node in a Binary Treefunction findHeight(root, x){ // Stores height of the Tree findHeightUtil(root, x); // Return the height return height;}// Driver Code// Binary Tree Formationvar root = newNode(5);root.left = newNode(10);root.right = newNode(15);root.left.left = newNode(20);root.left.right = newNode(25);root.left.right.right = newNode(45);root.right.left = newNode(30);root.right.right = newNode(35);var k = 25;// Function call to find the// depth of a given nodedocument.write("Depth: " + findDepth(root, k)+"<br>");// Function call to find the// height of a given nodedocument.write("Height: " + findHeight(root, k));</script> |
Output
Depth: 2 Height: 1
Time Complexity: O(N)
Auxiliary Space: O(1)
Another Approach(Using Level Order Traversal): Simple and Easy to understand
Follow the below steps to solve the above problem:
1) Initialize height and depth variable with -1;
2) Initialize a queue and a level variable with 0 and push the root in the queue.
3) Perform level order traversal and if value of frontNode is equal to the target(K) node then value of depth will be equal to the level value and continue traversing.
4) After completion we can calculate the value of height using height = level – depth – 1;
5) Print the value of height and depth variable.
Below is the implementation of above approach:
C++
// C++ Program to solve the above problem#include<bits/stdc++.h>using namespace std;struct Node{ int data; Node* left; Node* right; Node(int item){ data = item; left = right = NULL; }};Node* newNode(int value){ return new Node(value);}void findDepthAndHeight(Node* root, int k){ if(root == NULL) return; int depth = -1; int height = -1; queue<Node*> q; q.push(root); int level = 0; while(!q.empty()){ int n = q.size(); for(int i = 0; i<n; i++){ Node* frontNode = q.front(); q.pop(); if(frontNode->data == k) depth = level; if(frontNode->left) q.push(frontNode->left); if(frontNode->right) q.push(frontNode->right); } level++; } height = level - depth - 1; cout<<"Depth : "<<depth<<endl; cout<<"Height : "<<height<<endl;}int main(){ // Binary Tree Formation Node* root = newNode(5); root->left = newNode(10); root->right = newNode(15); root->left->left = newNode(20); root->left->right = newNode(25); root->left->right->right = newNode(45); root->right->left = newNode(30); root->right->right = newNode(35); int k = 25; findDepthAndHeight(root, k); return 0;} |
Java
// Java program to solve the above problemimport java.util.LinkedList;import java.util.Queue;class Node { int data; Node left; Node right; Node(int item) { data = item; left = right = null; }}public class GFG { static Node newNode(int value) { return new Node(value); } static void findDepthAndHeight(Node root, int k) { if (root == null) return; int depth = -1; int height = -1; Queue<Node> q = new LinkedList<>(); q.add(root); int level = 0; while (!q.isEmpty()) { int n = q.size(); for (int i = 0; i < n; i++) { Node frontNode = q.poll(); if (frontNode.data == k) depth = level; if (frontNode.left != null) q.add(frontNode.left); if (frontNode.right != null) q.add(frontNode.right); } level++; } height = level - depth - 1; System.out.println("Depth : " + depth); System.out.println("Height : " + height); } public static void main(String[] args) { // Binary Tree Formation Node root = newNode(5); root.left = newNode(10); root.right = newNode(15); root.left.left = newNode(20); root.left.right = newNode(25); root.left.right.right = newNode(45); root.right.left = newNode(30); root.right.right = newNode(35); int k = 25; findDepthAndHeight(root, k); }} |
Python3
class Node: def __init__(self, data): self.data = data self.left = None self.right = Nonedef new_node(value): return Node(value)def find_depth_and_height(root, k): if root is None: return depth = -1 height = -1 queue = [] queue.append(root) level = 0 while queue: n = len(queue) for i in range(n): front_node = queue.pop(0) if front_node.data == k: depth = level if front_node.left: queue.append(front_node.left) if front_node.right: queue.append(front_node.right) level += 1 height = level - depth - 1 print("Depth:", depth) print("Height:", height)def main(): # Binary Tree Formation root = new_node(5) root.left = new_node(10) root.right = new_node(15) root.left.left = new_node(20) root.left.right = new_node(25) root.left.right.right = new_node(45) root.right.left = new_node(30) root.right.right = new_node(35) k = 25 find_depth_and_height(root, k)if __name__ == "__main__": main() |
C#
using System;using System.Collections.Generic;public class Node { public int data; public Node left; public Node right; public Node(int item) { data = item; left = right = null; }}public class GFG { static Node newNode(int value) { return new Node(value); } static void findDepthAndHeight(Node root, int k) { if (root == null) return; int depth = -1; int height = -1; Queue<Node> q = new Queue<Node>(); q.Enqueue(root); int level = 0; while (q.Count != 0) { int n = q.Count; for (int i = 0; i < n; i++) { Node frontNode = q.Dequeue(); if (frontNode.data == k) depth = level; if (frontNode.left != null) q.Enqueue(frontNode.left); if (frontNode.right != null) q.Enqueue(frontNode.right); } level++; } height = level - depth - 1; Console.WriteLine("Depth : " + depth); Console.WriteLine("Height : " + height); } public static void Main(string[] args) { // Binary Tree Formation Node root = newNode(5); root.left = newNode(10); root.right = newNode(15); root.left.left = newNode(20); root.left.right = newNode(25); root.left.right.right = newNode(45); root.right.left = newNode(30); root.right.right = newNode(35); int k = 25; findDepthAndHeight(root, k); }} |
Javascript
class TreeNode { constructor(value) { this.data = value; this.left = null; this.right = null; }}function findDepthAndHeight(root, k) { if (root === null) return; let depth = -1; let height = -1; const queue = []; queue.push(root); let level = 0; while (queue.length > 0) { const n = queue.length; for (let i = 0; i < n; i++) { const frontNode = queue.shift(); if (frontNode.data === k) depth = level; if (frontNode.left !== null) queue.push(frontNode.left); if (frontNode.right !== null) queue.push(frontNode.right); } level++; } height = level - depth - 1; console.log("Depth: " + depth); console.log("Height: " + height);}// Binary Tree Formationconst root = new TreeNode(5);root.left = new TreeNode(10);root.right = new TreeNode(15);root.left.left = new TreeNode(20);root.left.right = new TreeNode(25);root.left.right.right = new TreeNode(45);root.right.left = new TreeNode(30);root.right.right = new TreeNode(35);const k = 25;findDepthAndHeight(root, k); |
Output
Depth : 2 Height : 1
Time Complexity: O(N), where N is the number of nodes in a given binary tree.
Auxiliary Space: O(N) due to queue data structure.
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