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;
        }
    }
}