【LEETCODE】230-Kth Smallest Element in a BST

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.

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?

Hint:

  1. Try to utilize the property of a BST.
  2. What if you could modify the BST node's structure?
  3. The optimal runtime complexity is O(height of BST).

题意:

给一个二叉搜索树,写一个函数,找到第k小的元素


注意:

可以假设k一直是有效的,即1 ≤ k ≤ BST的元素个数总和


思考:

如果这个BST经常被修改进行插入删除等操作并且需要频繁地找到第k小的元素,你需要怎么优化函数?


提示:

1.利用BST的属性

2.如何可以修改BST的节点结构,该如何做

3.最优时间复杂度是O(height of BST)


参考:

http://bookshadow.com/weblog/2015/07/02/leetcode-kth-smallest-element-bst/

BST具有如下性质:
左子树中所有元素的值均小于根节点的值
右子树中所有元素的值均大于根节点的值
因此采用中序遍历(左 -> 根 -> 右)即可以递增顺序访问BST中的节点,从而得到第k小的元素,时间复杂度O(k)



Python

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

class Solution(object):
    def kthSmallest(self, root, k):
        """
        :type root: TreeNode
        :type k: int
        :rtype: int
        """
        
        stack=[]
        node=root
        
        while node:
            stack.append(node)
            node=node.left
        
        x=1
        
        while stack and x<=k:
            node=stack.pop()
            x+=1
            right=node.right
            
            while right:
                stack.append(right)
                right=right.left
        
        return node.val





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