给定二叉搜索树的根 root 、一个目标值 target 和一个整数 k ,返回BST中最接近目标的 k 个值。你可以按 任意顺序 返回答案。
题目 保证 该二叉搜索树中只会存在一种 k 个值集合最接近 target
示例 1:

输入: root = [4,2,5,1,3],目标值 = 3.714286,且 k = 2
输出: [4,3]
示例 2:
输入: root = [1], target = 0.000000, k = 1
输出: [1]
提示:
二叉树的节点总数为 n
1 <= k <= n <= 104
0 <= Node.val <= 109
-109 <= target <= 109
进阶:假设该二叉搜索树是平衡的,请问您是否能在小于 O(n)( n = total nodes )的时间复杂度内解决该问题呢?
来源:力扣(LeetCode)
链接:https://leetcode.cn/problems/closest-binary-search-tree-value-ii
Python3提交内容:
# 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]:
if not root:
retrun
res = []
self.inorder(root, target, k, res)
return res
def inorder(self, root, target, k, res):
if not root:
return
self.inorder(root.left, target, k, res)
if k > len(res):
res.append(root.val)
elif abs(res[0] - target) > abs(root.val - target):
res.pop(0)
res.append(root.val)
else:
return
self.inorder(root.right, target, k, res)