Kth Smallest Element in a BST - LeetCode

题意

Given a binary search tree, write a function kthSmallest to find the kth smallest element in it.

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?

Constraints:

• The number of elements of the BST is between 1 to 10^4.
• You may assume k is always valid, 1 ≤ k ≤ BST's total elements.

思路

1. 递归：先找左子树，再找右子树。本质也是中序遍历。
2. 也可以中序遍历，直到第k个

题解

# 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: TreeNode, k: int) -> int:
        return self.dfs(root, k)[1].val
    
    def dfs(self, root, k):
        if root is None: return 0, None
        left = self.dfs(root.left, k)
        if left[1]: return left
        if left[0] == k - 1:
            return k, root
        right = self.dfs(root.right, k - left[0] - 1)
        if right[1]: return right
        return left[0] + right[0] + 1, None

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