230. Kth Smallest Element in a BST

Problem

Given the root of a binary search tree, and an integer k, return the kᵗʰ smallest value (1-indexed) of all the values of the nodes in the tree.

https://leetcode.com/problems/kth-smallest-element-in-a-bst/

Example 1:

Input: root = [3,1,4,null,2], k = 1
Output: 1

Example 2:

Input: root = [5,3,6,2,4,null,null,1], k = 3
Output: 3

Constraints:

The number of nodes in the tree is n.
1 <= k <= n <= 10⁴
0 <= Node.val <= 10⁴

Follow up: If the BST is modified often (i.e., we can do insert and delete operations) and you need to find the kth smallest frequently, how would you optimize?

Test Cases

python
# 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 kthSmallest(self, root: Optional[TreeNode], k: int) -> int:

solution_test.py下载完整源码
py
import pytest

import os
import sys
sys.path.append(os.path.dirname(os.path.dirname(__file__)))
from _utils.binary_tree import build_tree
from solution import Solution

null = None


@pytest.mark.parametrize('root, k, expected', [
    ([3,1,4,null,2], 1, 1),
    ([5,3,6,2,4,null,null,1], 3, 3),
])
class Test:
    def test_solution(self, root, k, expected):
        sol = Solution()
        assert sol.kthSmallest(build_tree(root), k) == expected

Thoughts

直接按照中序（in-order，LNR）遍历 BST 二叉树。访问到的第 k 个节点，其值就是第 k 小的数。

时间复杂度 O(n)。虽然找到第 k 个节点就结束，但中序遍历的话，即便是找到值最小的节点，最坏情况也可能需要先路过所有的其他节点。

所以频繁增删的 BST，可能会变得不平衡导致需要花很长时间才能找到第 k 小的节点，可以对 BST 做平衡调节。

Code

solution.py下载完整源码
py
from typing import Optional


# 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 kthSmallest(self, root: Optional[TreeNode], k: int) -> int:
        i = 0
        stack = []
        while root or stack:
            if root:
                stack.append(root)
                root = root.left
            else:
                root = stack.pop()
                if (i := i + 1) == k:
                    return root.val
                root = root.right

#Problem

#Test Cases

#Thoughts

#Code

Problem

Test Cases

Thoughts

Code