LeetCode: Minimum Depth of Binary Tree

简单题，少数次过

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int dfs(TreeNode *root, int dep) {
        if (!root) return dep;
        if (root->left && root->right) return min(dfs(root->left, dep+1), dfs(root->right, dep+1));
        if (!root->left && root->right) return dfs(root->right, dep+1);
        if (root->left && !root->right) return dfs(root->left, dep+1);
        if (!root->left && !root->right) return dep+1;
    }
    int minDepth(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        return dfs(root, 0);
    }
};

 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。。

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