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

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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?

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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 = None

class 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;
    }
};