LeetCode:Kth Smallest Element in a BST

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.

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?

思路：BST中序遍历 得到有序序列 以下用的是递归 ，当然也可以用栈

解法一：看leetcode提交记录

/**
 * 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 {
private:
         int count=0;
         int result=0;
public:
    int kthSmallest(TreeNode* root, int k) {
        if(root==NULL) return 0;
          kthSmallest(root->left,k);
        if(count<k)
        {
        count++;
        if(count==k)
           result=root->val;         
        }
        if(root->right!=NULL)
          kthSmallest(root->right,k);
          return result;
        
    }
     
};

代码二：

/**
 * 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) {
         int count=0;
         int result=0;
         inorderTraversal(root,count,result,k);
         return result;
       
    }
    void inorderTraversal(TreeNode *node,int &count,int &result,int &k)
    {
        if(node==NULL) return;
        inorderTraversal(node->left,count,result,k);
        
        if(count<k) 
        {count++;
        if(count==k) 
            {
                result=node->val;
                return ;
            }
        }
         inorderTraversal(node->right,count,result,k);
    }
    
   
    
};