Leetcode #2476: Closest Nodes Queries in a Binary Search Tree
In this guide, we solve Leetcode #2476 Closest Nodes Queries in a Binary Search Tree in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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].
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Tree, Depth-First Search, Binary Search Tree, Array, Binary Search, Binary Tree
Intuition
The problem structure suggests a monotonic decision, which makes binary search a natural fit.
By halving the search space each step, we reach the answer efficiently.
Approach
Search either directly on a sorted array or on the answer space using a check function.
Each check is fast, and the logarithmic search keeps the overall runtime low.
Steps:
- Define the search bounds.
- Check the mid point condition.
- Narrow the bounds until convergence.
Example
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].
Python Solution
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def closestNodes(
self, root: Optional[TreeNode], queries: List[int]
) -> List[List[int]]:
def dfs(root: Optional[TreeNode]):
if root is None:
return
dfs(root.left)
nums.append(root.val)
dfs(root.right)
nums = []
dfs(root)
ans = []
for x in queries:
i = bisect_left(nums, x + 1) - 1
j = bisect_left(nums, x)
mi = nums[i] if 0 <= i < len(nums) else -1
mx = nums[j] if 0 <= j < len(nums) else -1
ans.append([mi, mx])
return ans
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.