[LeetCode] Second Minimum Node In a Binary Tree

Posted 2020-10-06 immjc

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.

求二叉树中的第二小的节点：首先遍历二叉树后排序。然后在数组中找出第二小的元素即可。如果不存在第二小的元素，则返回-1

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> res;
    int findSecondMinimumValue(TreeNode* root) {
        inorder(root);
        sort(res.begin(), res.end());
        int smaller = -1;
        for (int i = 0; i != res.size() - 1; i++) {
            if (res[i + 1] - res[i] > 0) {
                smaller = res[i+1];
                break;
            }
        }
        return smaller;
    }
    vector<int> inorder(TreeNode* root) {
        if (root == nullptr)
            return res;
        inorder(root->left);
        res.push_back(root->val);
        inorder(root->right);
        return res;
    }
};
// 3 ms