leetcode 230 二叉搜索树

没想到，二叉搜索树的中序遍历就是按照顺序的，长知识了。所以通过中序遍历二叉搜索树，就可以获得按大小排列的数组，就此获得第K小的结果是很自然的，贴代码

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 8  *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 9  *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
10  * };
11  */
12 class Solution {
13 public:
14     int kthSmallest(TreeNode* root, int k) 
15     {
16         dfs(root);
17         return nodes[k-1];
18     }
19 private:
20     int i = 0;
21     vector<int> nodes;
22     void dfs(TreeNode* root)
23     {
24         if(!root)
25         {
26             return;
27         }
28         dfs(root->left);
29         nodes.push_back(root->val);
30         dfs(root->right);
31     }
32 };

标签：right,TreeNode,val,int,root,二叉,230,leetcode,left
来源： https://www.cnblogs.com/zhaohhhh/p/15162239.html