Description
Given a binary search tree, write a function kthSmallest to find the kth smallest element in it.
Note: You may assume k is always valid, 1 ≤ k ≤ BST’s total elements.
Example 1:
Input: root = [3,1,4,null,2], k = 1
3
/ \
1 4
\
2
Output: 1
Example 2:
Input: root = [5,3,6,2,4,null,null,1], k = 3
5
/ \
3 6
/ \
2 4
/
1
Output: 3
Follow up: What if the BST is modified (insert/delete operations) often and you need to find the kth smallest frequently? How would you optimize the kthSmallest routine?
Solutions
1. Inorder
# Time: O(logn)
# Space: O(n)
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def kthSmallest(self, root: TreeNode, k: int) -> int:
inorder = []
node = root
stack = []
while node or stack:
while node:
stack.append(node)
node = node.left
node = stack.pop()
inorder.append(node.val)
if len(inorder) >= k:
return inorder[-1]
node = node.right
return -1
# 91/91 cases passed (48 ms)
# Your runtime beats 80.57 % of python3 submissions
# Your memory usage beats 98.18 % of python3 submissions (16.8 MB)