Kth Smallest Element in a BST —— Leetcode

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).

用的BST性质就是，左侧都比root小，右侧都比root大。

如果每次用递归来计算左子树或右子树的结点个数，类似于二分法，但每次计算左子树节点个数会有很多重复开销；如果按照提示，改变TreeNode的存储结构，再添加一个leftCount属性，那么就比较容易计算出结果了。

下面提供了一种用stack来保存节点的方法，思想是将节点入栈，保证从最小节点开始遍历直到找到第k个，时间复杂度是O(k)；

/** * 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) {        stack<TreeNode *> st;        TreeNode* p = root;        while(p || !st.empty()) {            while(p) {                st.push(p);                p = p->left;            }            p = st.top();            if(--k == 0) {                return p->val;            }                        st.pop();            p = p->right;        }    }};