Given a binary tree, find its minimum depth. The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.
For example, minimum depth of below Binary Tree is 2.
Note that the path must end on a leaf node. For example, the minimum depth of below Binary Tree is also 2.
10
/
5
Method 1: The idea is to traverse the given Binary Tree. For every node, check if it is a leaf node. If yes, then return 1. If not leaf node then if the left subtree is NULL, then recur for the right subtree. And if the right subtree is NULL, then recur for the left subtree. If both left and right subtrees are not NULL, then take the minimum of two depths.
Below is implementation of the above idea.
C++
// C++ program to find minimum depth of a given Binary Tree#include<bits/stdc++.h>using namespace std;// A BT Nodestruct Node{ int data; struct Node* left, *right;};int minDepth(Node *root){ // Corner case. Should never be hit unless the code is // called on root = NULL if (root == NULL) return 0; // Base case : Leaf Node. This accounts for height = 1. if (root->left == NULL && root->right == NULL) return 1; int l = INT_MAX, r = INT_MAX; // If left subtree is not NULL, recur for left subtree if (root->left) l = minDepth(root->left); // If right subtree is not NULL, recur for right subtree if (root->right) r = minDepth(root->right); //height will be minimum of left and right height +1 return min(l , r) + 1;}// Utility function to create new NodeNode *newNode(int data){ Node *temp = new Node; temp->data = data; temp->left = temp->right = NULL; return (temp);}// Driver programint main(){ // Let us construct the Tree shown in the above figure Node *root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(4); root->left->right = newNode(5); cout <<"The minimum depth of binary tree is : "<< minDepth(root); return 0;} |
C
// C program to find minimum depth of a given Binary Tree#include <limits.h>#include <stdio.h>#include <stdlib.h>// A BT Nodetypedef struct Node { int data; struct Node *left, *right;} Node;int min(int num1, int num2){ return (num1 > num2) ? num2 : num1;}int minDepth(Node* root){ // Corner case. Should never be hit unless the code is // called on root = NULL if (root == NULL) return 0; // Base case : Leaf Node. This accounts for height = 1. if (root->left == NULL && root->right == NULL) return 1; int l = INT_MAX; int r = INT_MIN; // If left subtree is not NULL, recur for left subtree if (root->left) l = minDepth(root->left); // If right subtree is not NULL, recur for right subtree if (root->right) r = minDepth(root->right); // height will be minimum of left and right height +1 return min(l, r) + 1;}// Utility function to create new NodeNode* newNode(int data){ Node* temp = (Node*)malloc(sizeof(Node)); temp->data = data; temp->left = temp->right = NULL; return (temp);}// Driver programint main(){ // Let us construct the Tree shown in the above figure Node* root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(4); root->left->right = newNode(5); printf("The minimum depth of binary tree is : %d", minDepth(root)); return 0;}// This code is contributed by aditya kumar (adityakumar129) |
Java
/* Java implementation to find minimum depth of a given Binary tree *//* Class containing left and right child of currentnode and key value*/class Node{ int data; Node left, right; public Node(int item) { data = item; left = right = null; }}public class BinaryTree{ //Root of the Binary Tree Node root; int minimumDepth() { return minimumDepth(root); } /* Function to calculate the minimum depth of the tree */ int minimumDepth(Node root) { // Corner case. Should never be hit unless the code is // called on root = NULL if (root == null) return 0; // Base case : Leaf Node. This accounts for height = 1. if (root.left == null && root.right == null) return 1; // If left subtree is NULL, recur for right subtree if (root.left == null) return minimumDepth(root.right) + 1; // If right subtree is NULL, recur for left subtree if (root.right == null) return minimumDepth(root.left) + 1; return Math.min(minimumDepth(root.left), minimumDepth(root.right)) + 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("The minimum depth of "+ "binary tree is : " + tree.minimumDepth()); }} |
Python3
# Python program to find minimum depth of a given Binary Tree# Tree nodeclass Node: def __init__(self , key): self.data = key self.left = None self.right = Nonedef minDepth(root): # Corner Case.Should never be hit unless the code is # called on root = NULL if root is None: return 0 # Base Case : Leaf node.This accounts for height = 1 if root.left is None and root.right is None: return 1 # If left subtree is Null, recur for right subtree if root.left is None: return minDepth(root.right)+1 # If right subtree is Null , recur for left subtree if root.right is None: return minDepth(root.left) +1 return min(minDepth(root.left), minDepth(root.right))+1# Driver Program root = Node(1)root.left = Node(2)root.right = Node(3)root.left.left = Node(4)root.left.right = Node(5)print (minDepth(root))# This code is contributed by Nikhil Kumar Singh(nickzuck_007) |
C#
using System;/* C# implementation to find minimum depth of a given Binary tree *//* 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; }}public class BinaryTree{ //Root of the Binary Tree public Node root; public virtual int minimumDepth() { return minimumDepth(root); } /* Function to calculate the minimum depth of the tree */ public virtual int minimumDepth(Node root) { // Corner case. Should never be hit unless the code is // called on root = NULL if (root == null) { return 0; } // Base case : Leaf Node. This accounts for height = 1. if (root.left == null && root.right == null) { return 1; } // If left subtree is NULL, recur for right subtree if (root.left == null) { return minimumDepth(root.right) + 1; } // If right subtree is NULL, recur for left subtree if (root.right == null) { return minimumDepth(root.left) + 1; } return Math.Min(minimumDepth(root.left), minimumDepth(root.right)) + 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); Console.WriteLine("The minimum depth of binary tree is : " + tree.minimumDepth()); }} // This code is contributed by Shrikant13 |
Javascript
<script>/* javascript implementation to find minimum depth of a given Binary tree *//* Class containing left and right child of currentnode and key value*/class Node { constructor(item) { this.data = item; this.left = this.right = null; }} // Root of the Binary Tree let root; function minimumDepth() { return minimumDepth(root); } /* Function to calculate the minimum depth of the tree */ function minimumDepth( root) { // Corner case. Should never be hit unless the code is // called on root = NULL if (root == null) return 0; // Base case : Leaf Node. This accounts for height = 1. if (root.left == null && root.right == null) return 1; // If left subtree is NULL, recur for right subtree if (root.left == null) return minimumDepth(root.right) + 1; // If right subtree is NULL, recur for left subtree if (root.right == null) return minimumDepth(root.left) + 1; return Math.min(minimumDepth(root.left), minimumDepth(root.right)) + 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("The minimum depth of " + "binary tree is : " + minimumDepth(root));// This code contributed by aashish1995 </script> |
Output
The minimum depth of binary tree is : 2
Time Complexity: O(n), as it traverses the tree only once.
Auxiliary Space: O(h), where h is the height of the tree, this space is due to the recursive call stack.
Method 2: The above method may end up with complete traversal of Binary Tree even when the topmost leaf is close to root. A Better Solution is to do Level Order Traversal. While doing traversal, returns depth of the first encountered leaf node.
Below is the implementation of this solution.
C++
// C++ program to find minimum depth of a given Binary Tree#include <bits/stdc++.h>using namespace std;// A Binary Tree Nodestruct Node { int data; struct Node *left, *right; Node(int d = 0) : data{ d } { }};// A queue item (Stores pointer to node and an integer)struct qItem { Node* node; int depth;};// Iterative method to find minimum depth of Binary Treeint minDepth(Node* root){ // Corner Case if (root == NULL) return 0; // Create an empty queue for level order traversal queue<qItem> q; // Enqueue Root and initialize depth as 1 qItem qi = { root, 1 }; q.push(qi); // Do level order traversal while (q.empty() == false) { // Remove the front queue item qi = q.front(); q.pop(); // Get details of the remove item Node* node = qi.node; int depth = qi.depth; // If this is the first leaf node seen so far // Then return its depth as answer if (node->left == NULL && node->right == NULL) return depth; // If left subtree is not NULL, add it to queue if (node->left != NULL) { qi.node = node->left; qi.depth = depth + 1; q.push(qi); } // If right subtree is not NULL, add it to queue if (node->right != NULL) { qi.node = node->right; qi.depth = depth + 1; q.push(qi); } } 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 = 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); cout << minDepth(root); return 0;} |
Java
// Java program to find minimum depth // of a given Binary Tree import java.util.*;public class GFG { // A binary Tree node static class Node { int data; Node left, right; }// A queue item (Stores pointer to // node and an integer) static class qItem { Node node; int depth; public qItem(Node node, int depth) { this.node = node; this.depth = depth; }} // Iterative method to find// minimum depth of Binary Tree static int minDepth(Node root) { // Corner Case if (root == null) return 0; // Create an empty queue for level order traversal Queue<qItem> q = new LinkedList<>(); // Enqueue Root and initialize depth as 1 qItem qi = new qItem(root, 1); q.add(qi); // Do level order traversal while (q.isEmpty() == false) { // Remove the front queue item qi = q.peek(); q.remove(); // Get details of the remove item Node node = qi.node; int depth = qi.depth; // If this is the first leaf node seen so far // Then return its depth as answer if (node.left == null && node.right == null) return depth; // If left subtree is not null, // add it to queue if (node.left != null) { qi.node = node.left; qi.depth = depth + 1; q.add(qi); } // If right subtree is not null, // add it to queue if (node.right != null) { qi.node = node.right; qi.depth = depth + 1; q.add(qi); } } return 0; } // Utility function to create a new tree Node static Node newNode(int data) { Node temp = new Node(); temp.data = data; temp.left = temp.right = null; return temp; } // Driver Codepublic static void main(String[] args){ // 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); System.out.println(minDepth(root));}} // This code is contributed by 29AjayKumar |
Python3
# Python program to find minimum depth of a given Binary Tree# A Binary Tree nodeclass Node: # Utility to create new node def __init__(self , data): self.data = data self.left = None self.right = Nonedef minDepth(root): # Corner Case if root is None: return 0 # Create an empty queue for level order traversal q = [] # Enqueue root and initialize depth as 1 q.append({'node': root , 'depth' : 1}) # Do level order traversal while(len(q)>0): # Remove the front queue item queueItem = q.pop(0) # Get details of the removed item node = queueItem['node'] depth = queueItem['depth'] # If this is the first leaf node seen so far # then return its depth as answer if node.left is None and node.right is None: return depth # If left subtree is not None, add it to queue if node.left is not None: q.append({'node' : node.left , 'depth' : depth+1}) # if right subtree is not None, add it to queue if node.right is not None: q.append({'node': node.right , 'depth' : depth+1})# Driver program to test above function# Lets construct a binary tree shown in above diagramroot = Node(1)root.left = Node(2)root.right = Node(3)root.left.left = Node(4)root.left.right = Node(5)print (minDepth(root))# This code is contributed by Nikhil Kumar Singh(nickzuck_007) |
C#
// C# program to find minimum depth // of a given Binary Tree using System;using System.Collections.Generic; class GFG { // A binary Tree node public class Node { public int data; public Node left, right; }// A queue item (Stores pointer to // node and an integer) public class qItem { public Node node; public int depth; public qItem(Node node, int depth) { this.node = node; this.depth = depth; }} // Iterative method to find// minimum depth of Binary Tree static int minDepth(Node root) { // Corner Case if (root == null) return 0; // Create an empty queue for // level order traversal Queue<qItem> q = new Queue<qItem>(); // Enqueue Root and initialize depth as 1 qItem qi = new qItem(root, 1); q.Enqueue(qi); // Do level order traversal while (q.Count != 0) { // Remove the front queue item qi = q.Peek(); q.Dequeue(); // Get details of the remove item Node node = qi.node; int depth = qi.depth; // If this is the first leaf node // seen so far. // Then return its depth as answer if (node.left == null && node.right == null) return depth; // If left subtree is not null, // add it to queue if (node.left != null) { qi.node = node.left; qi.depth = depth + 1; q.Enqueue(qi); } // If right subtree is not null, // add it to queue if (node.right != null) { qi.node = node.right; qi.depth = depth + 1; q.Enqueue(qi); } } return 0; } // Utility function to create a new tree Node static Node newNode(int data) { Node temp = new Node(); temp.data = data; temp.left = temp.right = null; return temp; } // Driver Codepublic static void Main(String[] args){ // 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); Console.WriteLine(minDepth(root));}}// This code is contributed by PrinciRaj1992 |
Javascript
<script>// Javascript program to find minimum depth// of a given Binary Treeclass Node{ // Utility function to create a new tree Node constructor(data) { this.data = data; this.left = this.right = null; }}class qItem{ constructor(node,depth) { this.node = node; this.depth = depth; }}function minDepth(root){ // Corner Case if (root == null) return 0; // Create an empty queue for // level order traversal let q = []; // Enqueue Root and initialize depth as 1 let qi = new qItem(root, 1); q.push(qi); // Do level order traversal while (q.length != 0) { // Remove the front queue item qi = q.shift(); // Get details of the remove item let node = qi.node; let depth = qi.depth; // If this is the first leaf node seen so far // Then return its depth as answer if (node.left == null && node.right == null) return depth; // If left subtree is not null, // add it to queue if (node.left != null) { qi.node = node.left; qi.depth = depth + 1; q.push(qi); } // If right subtree is not null, // add it to queue if (node.right != null) { qi.node = node.right; qi.depth = depth + 1; q.push(qi); } } return 0;}// Driver Code// Let us create binary tree shown// in above diagramlet 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(minDepth(root));// This code is contributed by rag2127</script> |
Output
2
Time Complexity: O(n), where n is the number of nodes in the given binary tree. This is due to the fact that we are visiting each node once.
Auxiliary Space: O(n), as we need to store the elements in a queue for level order traversal.
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