简单题，少数次过

1 /** 2  * Definition for binary tree 3  * struct TreeNode { 4  *     int val; 5  *     TreeNode *left; 6  *     TreeNode *right; 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8  * }; 9  */10 class Solution {11 public:12     int dfs(TreeNode *root, int dep) {13         if (!root) return dep;14         if (root->left && root->right) return min(dfs(root->left, dep+1), dfs(root->right, dep+1));15         if (!root->left && root->right) return dfs(root->right, dep+1);16         if (root->left && !root->right) return dfs(root->left, dep+1);17         if (!root->left && !root->right) return dep+1;18     }19     int minDepth(TreeNode *root) {20         // Start typing your C/C++ solution below21         // DO NOT write int main() function22         return dfs(root, 0);23     }24 };

 C#, 没有return 0会有编译错误，C#的要求比C++的严格，但是功能又不全。。太废。。

1 /** 2  * Definition for a binary tree node. 3  * public class TreeNode { 4  *     public int val; 5  *     public TreeNode left; 6  *     public TreeNode right; 7  *     public TreeNode(int x) { val = x; } 8  * } 9  */10 public class Solution {11     public int MinDepth(TreeNode root) {12         return dfs(root, 0);13     }14     public int dfs(TreeNode root, int dep) {15         if (root == null) return dep;16         if (root.left != null && root.right != null) return Math.Min(dfs(root.left, dep+1), dfs(root.right, dep+1));17         if (root.left == null && root.right != null) return dfs(root.right, dep+1);18         if (root.left != null && root.right == null) return dfs(root.left, dep+1);19         if (root.left == null && root.right == null) return dep + 1;20         return 0;21     }22 }