Leetcode 111

Leetcode 111, Minimum Depth of Binarytree, 难度easy

这道题跟112是一样的，只要稍微的改一下就可以了，也就是在每一次递归时将层数加一，最后对比左子树和右子树的层数，将小的层数返回。

class Solution {public:    int minDepth(TreeNode* root) {        int res = 0;        if(root == NULL) return 0;        return DFS(root, res);    }    int DFS(TreeNode* root, int res) {        res++;        int left = root -> left == NULL?  (root -> right == NULL? res: INT_MAX) : DFS(root -> left, res);        int right = root -> right == NULL? (root -> left == NULL? res: INT_MAX) : DFS(root -> right, res);        if(left < right) return left;        return right;    }};