671. Second Minimum Node In a Binary Tree - Easy

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

如果节点为空，或者没有子节点，返回-1

如果子节点的值和当前节点一样，递归求next candidate；如果不同，子节点的值当作candidate

如果左右节点的值都不是 -1，返回其中较小的节点值，否则返回其中不等于-1的节点值

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;
        }
    }
}