272. Closest Binary Search Tree Value II

题干：

Given a non-empty binary search tree and a target value, find k values in the BST that are closest to the target.

Note:

Given target value is a floating point.
You may assume k is always valid, that is: k ≤ total nodes.
You are guaranteed to have only one unique set of k values in the BST that are closest to the target.

解释：

寻找BST中最接近目标的K个数字

思考：

首先BST中序遍历为有序，先转换成有序数列。 然后按照和target的绝对值的差排序，找到返回最前面的K个，时间复杂度是nlogn 如果使用一个最大堆，时间复杂度是nlogk 还有更好的办法，就是从两头开始删除，每次删除距离较大的那个。这样只要On就行了。

答案：

class Solution:
    def closestKValues(self, root: TreeNode, target: float, k: int) -> List[int]:
        if not root: return []
        stack = []
        result = []
        while stack or root:
            while root:
                stack.append(root)
                root = root.left
            root = stack.pop()
            if len(result) < k: result.append(root.val)
            elif abs(result[0] - target) > abs(root.val - target):
                result.pop(0)
                result.append(root.val)
            else: break
            root = root.right
        return result

答案补充：