LeetCode 230. Kth Smallest Element in a BST--C++,Python解法--面试真题--找二叉树中第K小的元素

题目地址：Kth Smallest Element in a BST - LeetCode

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 = 13/ \1   4\2
Output: 1

Example 2:

Input: root = [5,3,6,2,4,null,null,1], k = 35/ \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?

这道题目是经典的查找二叉树中第K小的元素，标准做法就是中序遍历，找到第K个值。
递归的Python解法如下：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = Noneclass Solution:def kthSmallest(self, root: TreeNode, k: int) -> int:def inorder(r):return inorder(r.left) + [r.val] + inorder(r.right) if r else []return inorder(root)[k - 1]

迭代的C++解法如下：

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}* };*/
class Solution {public:int kthSmallest(TreeNode *root, int k) {stack<TreeNode *> s;TreeNode *current = root;TreeNode *prev = nullptr;while (current || !s.empty()) {while (current) {s.push(current);current = current->left;}current = s.top();s.pop();k--;if (k == 0) {break;}prev = current;current = current->right;}return current->val;}
};

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