lc2476.二叉搜索树最近节点查询

题目链接：

题解

题目中有坑，如果树退化成一个链，每次在树上搜索会超时。所以先把二叉搜索树转到一个数组（中序遍历），然后每次在数组上二分查找。

参考代码

/*
 * @lc app=leetcode.cn id=2476 lang=cpp
 *
 * [2476] 二叉搜索树最近节点查询
 */

// @lc code=start
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> closestNodes(TreeNode* root, vector<int>& queries) {
        vector<int> tmp;
        queue<TreeNode*> q;
        q.push(root);
        while(!q.empty()) {
            TreeNode* node = q.front();
            q.pop();
            tmp.push_back(node->val);
            if(node->left) q.push(node->left);
            if(node->right) q.push(node->right);
        }

        sort(tmp.begin(), tmp.end());
        vector<vector<int>> ans;
        for(int i = 0; i < queries.size(); i++) {
            int left = -1;
            int right = -1;
            int k = lower_bound(tmp.begin(), tmp.end(), queries[i]) - tmp.begin();
            if(k!=tmp.size() && tmp[k] == queries[i]) {
                left = queries[i];
                right = queries[i];
            }
            else if (k == 0) {right = tmp[k];}
            else if (k == tmp.size()) {left = tmp[k-1];}
            else {
                left = tmp[k-1];
                right = tmp[k];
            }


            ans.push_back({left, right});
        }
        return ans;
    }
};
// @lc code=end