[leetcode][BST] 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?

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).

基本解法：O(n)，其中n为BST的节点个数

/** * 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 (NULL == root || k < 1) return -1;//非法输入，应抛出异常的stack<TreeNode *> sta;TreeNode *p = root;int cnt = 0;while (p != NULL || !sta.empty()){while (p != NULL){//一路向左sta.push(p);p = p->left;}//visitTreeNode *q = sta.top();sta.pop();if (++cnt == k) return q->val;p = q->right;}}};

优化解法：

思路：在节点数据结构中增加一个表示在BST中该节点左边的节点个数的int型成员lCnt

插入：在查找节点插入位置的过程中，如果该节点应该放在当前节点的左边，则当前节点的lCnt加1，找到该节点的合适位置时，该节点的lCnt为0

删除：在查找删除节点的过程中，如果该节点值小于当前节点值，则当前节点的lCnt减1

查找：如果k小于当前节点的lCnt，到当前节点的左边继续查找；否则到当前节点的右边继续查找k=k-cur->lCnt-1

实现：待续