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.

ref: https://leetcode.com/problems/second-minimum-node-in-a-binary-tree/discuss/107158/Java-divide-and-conquer-solution

如果節點為空，或者沒有子節點，返回-1

如果子節點的值和當前節點一樣，遞迴求next candidate；如果不同，子節點的值當作candidate

如果左右節點的值都不是 -1，返回其中較小的節點值，否則返回其中不等於-1的節點值

time: O(n), space: O(height)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 
 
*/
class Solution {
    public int findSecondMinimumValue(TreeNode root) {
        if(root == null) {
            return -1;
        }
        if(root.left == null && root.right == null) {
            return -1;
        }
        int left = root.left.val;
        int right = root.right.val;
        if(root.left.val == root.val) {
            left = findSecondMinimumValue(root.left);
        }
        if(root.right.val == root.val) {
            right = findSecondMinimumValue(root.right);
        }
        
        if(left != -1 && right != -1) {
            return Math.min(left, right);
        } else if(left != -1) {
            return left;
        } else {
            return right;
        }
    }
}