Leetcode #272: Closest Binary Search Tree Value II
In this guide, we solve Leetcode #272 Closest Binary Search Tree Value II 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
Given the root of a binary search tree, a target value, and an integer k, return the k values in the BST that are closest to the target. You may return the answer in any order.
Quick Facts
- Difficulty: Hard
- Premium: Yes
- Tags: Stack, Tree, Depth-First Search, Binary Search Tree, Two Pointers, Binary Tree, Heap (Priority Queue)
Intuition
The constraints hint that we can reason about two ends of the data at once, which is perfect for a two-pointer scan.
Moving one pointer at a time keeps the invariant intact and avoids nested loops.
Approach
Place pointers at the left and right ends and move them based on the comparison or target condition.
This yields a clean linear pass after any required sorting.
Steps:
- Set left and right pointers.
- Move a pointer based on the condition.
- Update the best answer while scanning.
Example
Input: root = [4,2,5,1,3], target = 3.714286, k = 2
Output: [4,3]
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 closestKValues(self, root: TreeNode, target: float, k: int) -> List[int]:
def dfs(root):
if root is None:
return
dfs(root.left)
if len(q) < k:
q.append(root.val)
else:
if abs(root.val - target) >= abs(q[0] - target):
return
q.popleft()
q.append(root.val)
dfs(root.right)
q = deque()
dfs(root)
return list(q)
Complexity
The time complexity is O(n) (after optional sort O(n log n)). The space complexity is O(1).
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.