Second Minimum Node In a Binary Tree

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.

 

这道题是找二叉树中的第二小的结点值，二叉树的限制为，对于任意一个结点，要么其没有子结点，要么就同时有两个子结点，而且父结点值是子结点值中较小的那个，两个子结点值可以相等。

根据该树的附加条件可知，根结点一定是最小的结点值first，只要找出第二小的值second即可，初始化为整型的最大值。然后对根结点调用递归函数，将first和second当作参数传进去即可。在递归函数中，如果当前结点为空，直接返回，若当前结点孩值不等于first，说明其肯定比first要大，如果比second小的话就更新second，然后对当前结点的左右子结点分别调用递归函数即可。

class Solution {
public:
    int findSecondMinimumValue(TreeNode* root) {
        return helper(root, root->val);
    }
    int helper(TreeNode* node, int first) {
        if (!node) return -1;
        if (node->val != first) return node->val;
        int left = helper(node->left, first), right = helper(node->right, first);
        return (left == -1 || right == -1) ? max(left, right) : min(left, right);
    }
};