【LeetCode】230. 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?

Hint:

1. Try to utilize the property of a BST.
2. What if you could modify the BST node's structure?
3. The optimal runtime complexity is O(height of BST).

class Solution {public:    int kthSmallest(TreeNode* root, int k) {                stack<TreeNode*> stack;        TreeNode * p=root;        int result;        int count=k;                while((!stack.empty() || p!=NULL)){                        if(p!=NULL){                stack.push(p);                p=p->left;            }else{                                count--;                if(count==0){                    break;                }                p=stack.top();                stack.pop();                p=p->right;                            }        }                result =stack.top()->val;                return result;    }};

思路心得：

这题我的做法其实就是为了中序遍历加了一个计数器。然后就能记录下来BST中的元素大小了。

但是这题的思考部分是要为结点添加子树的大小这个数据。

感觉和非递归遍历BST比谁快?

虽然如果事先就有子树的大小的话确实是O(树高)