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   7

Output: 5
Explanation: The smallest value is 2, the second smallest value is 5.

Example 2:

Input: 
    2
   /   2   2

Output: -1
Explanation: The smallest value is 2, but there isn‘t any second smallest value.

思路：

用set存儲遍歷的值。取第二個即可。但是感覺不是什麽好辦法。。。以為麽有利用樹的結構信息。。暫時先這樣。。

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

[leetcode-671-Second Minimum Node In a Binary Tree]