LeetCode 111. Minimum Depth of Binary Tree

原题链接在这里：https://leetcode.com/problems/minimum-depth-of-binary-tree/

题目：

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.

题解：

与Maximum Depth of Binary Tree相似。

不同是多了判断叶子节点，若是给出[1,2]这种情况，返回值应该是2而不是1, 即使root的右节点为空，但根据定义：The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node. 此时root.left不为null, 所以root不能被认为是leaf. 因此加了if(root.left == null) return right+1, 这里注意加一。

Time Complexity: O(n)最坏的情况每个节点走了一遍. Space: O(logn), recursion stack 是 tree 的高度是logn.

AC Java:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int minDepth(TreeNode root) {
        if(root == null){
            return 0;
        }
        
        int left = minDepth(root.left);
        int right = minDepth(root.right);
        
        if(root.left == null){
            return right+1;    
        }
        if(root.right == null){
            return left+1;
        }
        
        return Math.min(left, right)+1;
    }
}

AC C++: 

/**
 * 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;
        }

        int l = minDepth(root->left);
        int r = minDepth(root->right);
        if(!root->left){
            return r + 1;
        }

        if(!root->right){
            return l + 1;
        }

        return min(l, r) + 1;
    }
};