2476. Closest Nodes Queries in a Binary Search Tree

You are given the root of a binary search tree and an array queries of size n consisting of positive integers.

Find a 2D array answer of size n where answer[i] = [mini, maxi]:

• mini is the largest value in the tree that is smaller than or equal to queries[i]. If a such value does not exist, add -1 instead.
• maxi is the smallest value in the tree that is greater than or equal to queries[i]. If a such value does not exist, add -1 instead.

Return the array answer.

/**
 * 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 {
    void dfs(TreeNode* node, vector<int>& now) {
        if(!node) return;
        now.push_back(node->val);
        dfs(node->left, now);
        dfs(node->right, now);
    }
public:
    vector<vector<int>> closestNodes(TreeNode* root, vector<int>& queries) {
        vector<int> now;
        dfs(root, now);
        sort(begin(now), end(now));
        now.erase(unique(begin(now),end(now)), end(now));
        vector<vector<int>> res;
        for(auto q : queries) {
            auto lb = lower_bound(begin(now), end(now), q);
            int mi = -1, ma = -1;
            if(lb != end(now)) ma = *lb;
            if(lb == end(now)) mi = *prev(lb);
            else {
                if(*lb == q) mi = *lb;
                else if(lb != begin(now)) mi = *prev(lb);
            }
            res.push_back({mi,ma});
        }
        return res;
    }
};