Leetcode 230. Kth Smallest Element in a BST

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

 

Medium

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?

---

• 因为是binary search tree，所以是有序的，变成DFS / BST遍历树。
• DFS
  • Time complexity : O(N) to build a traversal.
  • Space complexity : O(N) to keep an inorder traversal.
• BFS
  • Time complexity : O(H+k), where HH is a tree height. This complexity is defined by the stack, which contains at least H + kH+k elements, since before starting to pop out one has to go down to a leaf. This results in O(logN+k) for the balanced tree and O(N+k) for completely unbalanced tree with all the nodes in the left subtree.
  • Space complexity : O(H+k), the same as for time complexity, O(N+k) in the worst case, and O(logN+k) in the average case.

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def kthSmallest1(self, root: TreeNode, k: int) -> int:
        def inorder(r):
            return inorder(r.left) + [r.val] + inorder(r.right) if r else [] # pythonic style
        
        return inorder(root)[k - 1]     
    
    def kthSmallest(self, root: TreeNode, k: int) -> int:
        stack = []
        
        while True:
            while root:
                stack.append(root)
                root = root.left
            
            root = stack.pop()
            k -= 1
            
            if not k:
                return root.val
            
            root = root.right