To solve the problem, you need to find the closest node values for each query in a Binary Search Tree. Efficient solutions typically involve converting the tree into a sorted array and applying binary search to quickly find the closest smaller and greater values for each query. This approach leverages binary search properties and tree traversal to optimize performance for large inputs.
Problem Statement
Given the root of a Binary Search Tree and a list of queries, your task is to return a 2D array where each entry corresponds to a query, and contains two integers: the largest node value smaller or equal to the query, and the smallest node value greater or equal to the query. If no such value exists, return -1 for the respective value.
You can assume that the tree has at least two nodes and that the values of the queries are within a given range. You must optimize your solution to handle up to 100,000 queries and trees with up to 100,000 nodes. The challenge here is balancing the binary search and tree traversal techniques to efficiently find the answers for each query.
Examples
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]]
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]]
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
Solution Approach
Convert Tree to Sorted Array
Start by converting the Binary Search Tree into a sorted array of node values. This allows each query to be answered in logarithmic time using binary search techniques to find the closest nodes. A depth-first search (DFS) can be used to collect the values of the tree in sorted order.
Binary Search for Closest Values
For each query, perform a binary search on the sorted array of tree node values. Find the largest number smaller or equal to the query, and the smallest number greater or equal to the query. This can be done by tracking indices in the sorted array using binary search.
Handle Edge Cases
Ensure that the binary search handles edge cases like when no smaller value exists for a query, or when no larger value exists. In such cases, return -1 for the missing values. This ensures that every query has a valid response, even in cases where one of the closest values doesn't exist.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | Depends on the final approach |
| Space | Depends on the final approach |
The time complexity of the solution depends on converting the tree to a sorted array, which takes O(n) where n is the number of nodes in the tree. For each query, binary search takes O(log n). Thus, the overall time complexity for processing q queries is O(n + q log n). The space complexity is O(n) to store the sorted node values.
What Interviewers Usually Probe
- The candidate efficiently handles large inputs by reducing the problem to binary search on a sorted array.
- Look for correct handling of edge cases, especially the absence of a larger or smaller value for a query.
- The candidate demonstrates a solid understanding of binary search tree traversal and its application to solve problems efficiently.
Common Pitfalls or Variants
Common pitfalls
- Failing to optimize the solution for large inputs, resulting in a slower approach for processing the tree or queries.
- Incorrectly handling edge cases like when no smaller or larger node value exists for a query.
- Not converting the tree into a sorted array, leading to inefficient queries and unnecessary repeated traversal of the tree.
Follow-up variants
- What if the queries are given in random order, should the approach be adjusted to handle this?
- Consider a version where the tree is dynamic and can be updated between queries. How would that affect the solution?
- What if you have multiple trees and need to process queries for each one separately?
How GhostInterview Helps
- GhostInterview helps by offering guidance on efficiently converting the tree to a sorted array and using binary search to answer queries quickly.
- It can highlight potential pitfalls, like missing edge cases or inefficient approaches, to ensure that the solution is optimized for large inputs.
- GhostInterview can assist in refining the solution approach and complexity analysis, ensuring that the candidate provides a complete and efficient solution.
Topic Pages
Related GhostInterview Pages
- LeetCode Interview Copilot - Use GhostInterview as a live solver when you want direct help with LeetCode-style coding questions.
- Coding Interview Assistant - See how GhostInterview supports array, string, linked list, graph, and tree interview workflows.
- How GhostInterview Works - Review the screenshot, reasoning, and answer flow before using the solver in a live interview.
FAQ
What is the best way to handle large inputs in this problem?
The best approach is to first convert the Binary Search Tree into a sorted array, allowing each query to be answered in O(log n) time using binary search.
How do I handle queries where no smaller or larger value exists in the tree?
In such cases, return -1 for the missing value. Ensure the binary search algorithm accounts for these edge cases.
Is there a way to optimize the tree traversal for larger trees?
Yes, by converting the tree into a sorted array using a depth-first traversal (DFS), you can avoid repeated tree traversal for each query.
What is the time complexity of this problem?
The time complexity is O(n + q log n), where n is the number of nodes in the tree and q is the number of queries.
How do I approach this problem if the tree can be updated between queries?
If the tree is dynamic, you will need to update the sorted array as nodes are added or removed, and adjust the binary search accordingly.
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