# 671.Second Minimum Node in a Binary Tree

671.Second Minimum Node in a Binary Tree 难度:Easy

> 给定一个非空特殊的二叉树，每个节点都是正数，并且每个节点的子节点数量只能为 2 或 0。如果一个节点有两个子节点的话，那么这个节点的值不大于它的子节点的值。

给出这样的一个二叉树，你需要输出所有节点中的第二小的值。如果第二小的值不存在的话，输出 -1 。

```
示例 1:

输入: 
    2
   / \
  2   5
     / \
    5   7

输出: 5
说明: 最小的值是 2 ，第二小的值是 5 。
示例 2:

输入: 
    2
   / \
  2   2

输出: -1
说明: 最小的值是 2, 但是不存在第二小的值。
```

二叉樹的遍歷，不過需要注意的是，需要計數幾個不同的數來判斷是否存在第二小的數。

```
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
int min_fir=INT_MAX;
int min_sec=min_fir;
int count=0;
void traversal(TreeNode* root)
{
    if(!root)  return;
        if(root->val < min_fir) {min_sec=min_fir; min_fir=root->val ; ++count;}
        else if(root->val > min_fir && root->val <= min_sec) { min_sec=root->val; ++count;}
       
        if(root->left )
        {
            traversal(root->left);
            traversal(root->right);
        }
}
public:
    int findSecondMinimumValue(TreeNode* root) {
        traversal(root);
       // cout<<min_fir<<"   "<< min_sec<<endl;
        //cout<<count<<endl;
        return count<2 ? -1 : min_sec;
        
    }
};
```

> 执行用时 :4 ms, 在所有 C++ 提交中击败了83.38%的用户\
> 内存消耗 :8.4 MB, 在所有 C++ 提交中击败了95.10%的用户


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