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.

Solution：Tree

思路：
for left and right sub-node, if their value is the same with the parent node value, need to using recursion to find the next candidate, otherwise use their node value as the candidate.
Time Complexity: O(N) Space Complexity: O(N) 递归缓存

Solution Code：

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 value same as root value, need to find the next candidate
    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;
    }
}