Given a non-empty special binary tree consisting of nodes with the non-negative value, where each node in this tree has exactly two or 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.

思路：

用set存储遍历的值。取第二个即可。但是感觉不是什么好办法。。。以为么有利用树的结构信息。。暂时先这样。。

void pret(TreeNode*root, set
     
      &s){    if (root == NULL)return;    s.insert(root->val);    pret(root->left, s);    pret(root->right, s);    }int findSecondMinimumValue(TreeNode* root){    if (root == NULL)return -1;    set
      
       s;    pret(root, s);    if (s.size() < 2)return -1;    auto it = s.begin();    it++;    return *it;}