Introduction

In this blog post, we will explore how to solve the LeetCode problem "2476" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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. Example 1: Input: root = [6,2,13,1,4,9,15,null,null,null,null,null,null,14], queries = [2,5,16] Output: [[2,2],[4,6],[15,-1]] Explanation: We answer the queries in the following way: - The largest number that is smaller or equal than 2 in the tree is 2, and the smallest number that is greater or equal than 2 is still 2. So the answer for the first query is [2,2]. - The largest number that is smaller or equal than 5 in the tree is 4, and the smallest number that is greater or equal than 5 is 6. So the answer for the second query is [4,6]. - The largest number that is smaller or equal than 16 in the tree is 15, and the smallest number that is greater or equal than 16 does not exist. So the answer for the third query is [15,-1]. Example 2: Input: root = [4,null,9], queries = [3] Output: [[-1,4]] Explanation: The largest number that is smaller or equal to 3 in the tree does not exist, and the smallest number that is greater or equal to 3 is 4. So the answer for the query is [-1,4]. Constraints: The number of nodes in the tree is in the range [2, 105]. 1 <= Node.val <= 106 n == queries.length 1 <= n <= 105 1 <= queries[i] <= 106

Explanation

Here's the breakdown of the solution:

• Iterative BST Traversal: Utilize iterative inorder traversal with modifications to efficiently search for the floor (largest value <= query) and ceiling (smallest value >= query) for each query. This avoids recursion overhead and provides finer control over the traversal path.

• Maintain Floor and Ceiling: During traversal, update the floor and ceiling values based on comparisons with the current node's value. This allows for finding the nearest values meeting the criteria without having to traverse the entire tree for each query.

• Time & Space Complexity: O(N + Q log N) time, where N is the number of nodes in the BST and Q is the number of queries. Space complexity is O(1) excluding the answer array. O(Q) if including the answer array.

Code

    class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def closest_values(root, queries):
    def find_closest(root, query):
        floor_val = -1
        ceil_val = -1
        curr = root

        while curr:
            if curr.val == query:
                floor_val = curr.val
                ceil_val = curr.val
                break
            elif curr.val < query:
                floor_val = curr.val
                curr = curr.right
            else:
                ceil_val = curr.val
                curr = curr.left
        return [floor_val, ceil_val]

    result = []
    for query in queries:
        result.append(find_closest(root, query))
    return result