【LectCode】230. Kth Smallest Element in a BST

题目：

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?
解题思路：题目要求数的第k小的值，我们知道二分搜索树是按照值的大小分布在父节点的两边的，即若左子树的节点数为n,则根为第n+1小的值；因此，我先写一个计算树的节点数的函数，然后先计算左子树的节点数leftsize，若k == leftsize+1,则返回跟节点的值；若k <= leftsize,则进入左子树的递归；若k >= leftsize+1,则进入右子树的递归，并把k的值修改为k - leftsize -1。

解答：

/**

 * 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) {

         if(root == NULL){

        return 0;
}
int leftsize = cal(root->left);
if(k == leftsize +1){
return root->val;
}
else if(k <= leftsize){
return kthSmallest(root->left,k);
}
else{
return kthSmallest(root->right,k - leftsize -1);
}

    }

     int cal(TreeNode* root){

        if(root == NULL){

        return 0;
}
else{
return 1 + cal(root->left) + cal(root->right);
}
}

};