LeetCode : 671. Second Minimum Node In a Binary Tree 二叉树中次小节点

桃扇骨 2021-06-24 16:10 271阅读 0赞

**试题**  
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.  
**代码：**  
由于父节点肯定大于等于子节点，那么第一小肯定是根节点；这时由于子节点大于等于根节点，如果是等于的话我们需要继续找。

/**
     * 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 leftVal = root.left.val;
            if(leftVal==root.val)
                leftVal = findSecondMinimumValue(root.left);
            int rightVal = root.right.val;
            if(rightVal==root.val)
                rightVal = findSecondMinimumValue(root.right);
            if(leftVal!=-1 && rightVal!=-1){
                return Math.min(leftVal, rightVal);
            }else if(leftVal!=-1){
                return leftVal;
            }else if(rightVal!=-1){
                return rightVal;
            }else{
                return -1;
            }
            
        }
    }