leetcode 671. Second Minimum Node In a Binary Tree

原题：

Given a non-empty special binary tree consisting of nodes with the non-negative value, where each node in this tree has exactly twoor zero sub-node. If the node has two sub-nodes, then this node's value is the smaller value among its two sub-nodes.

Given such a binary tree, you need to output the second minimum value in the set made of all the nodes' value in the whole tree.

If no such second minimum value exists, output -1 instead.

Example 1:

Input:     2   / \  2   5     / \    5   7Output: 5Explanation: The smallest value is 2, the second smallest value is 5.

Example 2:

Input:     2   / \  2   2Output: -1Explanation: The smallest value is 2, but there isn't any second smallest value.

代码思路，还是用递归，找到第二小的数。只要找到左右两个节点中，第一个不是2的数，比较谁比较小就可以了。

代码如下：

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     struct TreeNode *left; *     struct TreeNode *right; * }; */int findSecondMinimumValue(struct TreeNode* root) {    int findsecond(struct TreeNode*);    if(root==NULL)        return -1;    int* second;    second=(int*)malloc(sizeof(int));    *second=root->val;    *second=findsecond(root);    if(INT_MAX==*second)        return -1;    return *second;}int findsecond(struct TreeNode* root){    if(root->right==NULL)        return INT_MAX;    int right=root->val;    int left=root->val;    if(root->right->val!=root->val)    {        right=root->right->val;    }    else    {        right=findsecond(root->right);    }    if(root->left->val!=root->val)    {        left=root->left->val;    }    else    {        left=findsecond(root->left);    }    //printf("A%d,%d",left,right);    return left<right?left:right;}

哦，没关系，哦，别在意。

疯狂到底，孤独至极。