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lletcode 230. 二叉搜索树中第K小的元素，链接：https://leetcode.cn/problems/kth-smallest-element-in-a-bst
题目描述
给定一个二叉搜索树的根节点 root ，和一个整数 k ，请你设计一个算法查找其中第 k 个最小元素（从 1 开始计数）。
解法一
直接dfs中序遍历，代码如下：

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode() : val(0), left(nullptr), right(nullptr) {}*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}* };*/
class Solution {
public:vector<int> res;int kthSmallest(TreeNode* root, int k) {dfs(root);return res[k - 1];}void dfs(TreeNode* root){if(!root)return;dfs(root->left);res.push_back(root->val);dfs(root->right);}
};

解法二

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode() : val(0), left(nullptr), right(nullptr) {}*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}* };*/
class Solution {
public:int kthSmallest(TreeNode* root, int k) {int res = 0;if(root == nullptr || k < 1) {return -1;}int numOfLeftNodes = getNumOfNodes(root->left);int numOfRightNodes = getNumOfNodes(root->right);if (numOfLeftNodes == k - 1) {return root->val;} else if (numOfLeftNodes >= k) {return kthSmallest(root->left, k);} else {return kthSmallest(root->right, k - numOfLeftNodes - 1);}}private:int getNumOfNodes(TreeNode* root) {if (!root) {return 0;}return getNumOfNodes(root->left) + getNumOfNodes(root->right) + 1;}
};