111. Minimum Depth of Binary Tree

Easy

Problem Description

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.

Note: A leaf is a node with no children.

Examples

Example 1:
Input: root = [3,9,20,null,null,15,7]
Output: 2

Example 2:
Input: root = [2,null,3,null,4,null,5,null,6]
Output: 5
Jump to Solution: Python Java C++ JavaScript C#

Python Solution


# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
from collections import deque

def minDepth(root: Optional[TreeNode]) -> int:
    if not root:
        return 0
    
    # Use BFS for shortest path
    queue = deque([(root, 1)])
    
    while queue:
        node, depth = queue.popleft()
        
        # If we find a leaf node, this is the minimum depth
        if not node.left and not node.right:
            return depth
        
        if node.left:
            queue.append((node.left, depth + 1))
        if node.right:
            queue.append((node.right, depth + 1))
    
    return 0  # Should never reach here for valid binary tree

Java Solution


/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int minDepth(TreeNode root) {
        if (root == null) {
            return 0;
        }
        
        // Use BFS with queue of node-depth pairs
        Queue> queue = new LinkedList<>();
        queue.offer(new Pair<>(root, 1));
        
        while (!queue.isEmpty()) {
            Pair pair = queue.poll();
            TreeNode node = pair.getKey();
            int depth = pair.getValue();
            
            // If we find a leaf node, this is the minimum depth
            if (node.left == null && node.right == null) {
                return depth;
            }
            
            if (node.left != null) {
                queue.offer(new Pair<>(node.left, depth + 1));
            }
            if (node.right != null) {
                queue.offer(new Pair<>(node.right, depth + 1));
            }
        }
        
        return 0;  // Should never reach here for valid binary tree
    }
}

C++ Solution


/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int minDepth(TreeNode* root) {
        if (!root) {
            return 0;
        }
        
        // Use BFS with queue of node-depth pairs
        queue> q;
        q.push({root, 1});
        
        while (!q.empty()) {
            auto [node, depth] = q.front();
            q.pop();
            
            // If we find a leaf node, this is the minimum depth
            if (!node->left && !node->right) {
                return depth;
            }
            
            if (node->left) {
                q.push({node->left, depth + 1});
            }
            if (node->right) {
                q.push({node->right, depth + 1});
            }
        }
        
        return 0;  // Should never reach here for valid binary tree
    }
};

JavaScript Solution


/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number}
 */
var minDepth = function(root) {
    if (!root) {
        return 0;
    }
    
    // Use BFS with queue of node-depth pairs
    const queue = [[root, 1]];
    
    while (queue.length) {
        const [node, depth] = queue.shift();
        
        // If we find a leaf node, this is the minimum depth
        if (!node.left && !node.right) {
            return depth;
        }
        
        if (node.left) {
            queue.push([node.left, depth + 1]);
        }
        if (node.right) {
            queue.push([node.right, depth + 1]);
        }
    }
    
    return 0;  // Should never reach here for valid binary tree
};

C# Solution


/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    public int MinDepth(TreeNode root) {
        if (root == null) {
            return 0;
        }
        
        // Use BFS with queue of node-depth pairs
        var queue = new Queue<(TreeNode node, int depth)>();
        queue.Enqueue((root, 1));
        
        while (queue.Count > 0) {
            var (node, depth) = queue.Dequeue();
            
            // If we find a leaf node, this is the minimum depth
            if (node.left == null && node.right == null) {
                return depth;
            }
            
            if (node.left != null) {
                queue.Enqueue((node.left, depth + 1));
            }
            if (node.right != null) {
                queue.Enqueue((node.right, depth + 1));
            }
        }
        
        return 0;  // Should never reach here for valid binary tree
    }
}

Complexity Analysis

Solution Explanation

This solution uses Breadth-First Search (BFS) to find the minimum depth:

Key points: