LeetCode 671. Second Minimum Node In a Binary Tree

原题链接在这里：https://leetcode.com/problems/second-minimum-node-in-a-binary-tree/

题目：

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.

题解：

If current root or root.left == null, then return -1.

Otherwise, check root.left.val and root.right.val, if they != root.val, it must be larger than root.val, could be candidate.

Otherwise, continue dfs on the side == root.val. 

If both sides != -1, return min, otherwise, return max.

Time Complexity: O(n). Space: O(logn).

AC Java:

/**
 * 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 || root.left == null){
            return -1;
        }
        
        int l = root.left.val;
        int r = root.right.val;
        if(l == root.val){
            l = findSecondMinimumValue(root.left);
        }
        
        if(r == root.val){
            r = findSecondMinimumValue(root.right);
        }
        
        if(l > 0 && r > 0){
            return Math.min(l, r);
        }else if(l > 0){
            return l;
        }else if(r > 0){
            return r;
        }else{
            return -1;
        }
    }
}