LeetCode: 230. Kth Smallest Element in a BST

题目

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

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.

解法

中序遍历生成数组，找出数组的第 k 个元素

通过 BST 的特性可以知道，中序遍历 BST 生成的所有节点的值组成的数组会是一个从小到大不包含重复值的有序数组，因此第 k 小的元素就是数组的第 k 个元素。

中序遍历的 Python 代码类似这样:

def inorder(root):
    if root is None:
        return
    inorder(root.left)

    # print(root)

    inorder(root.right)

这个方法的 Python 代码类似下面这样:

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):
    def kthSmallest(self, root, k):
        self.node_vals = []
        self.inorder(root)
        return self.node_vals[k-1]

    def inorder(self, root):
        if root is None:
            return
        self.inorder(root.left)

        self.node_vals.append(root.val)

        self.inorder(root.right)

中序遍历直接找到相应的元素

上面的方法中的数组其实可以不需要生成，直接在中序遍历的过程中找到相应的元素即可。

相应的 Python 代码类似下面这个:

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):
    def kthSmallest(self, root, k):
        self.number = 0
        self.node_val = None
        self.inorder(root, k)
        return self.node_val

    def inorder(self, root, k):
        if root is None:
            return

        self.inorder(root.left, k)

        self.number += 1
        if self.number == k:
            self.node_val = root.val
            return

        self.inorder(root.right, k)

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