leetcode230:Kth Smallest Element in a BST

Description:
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：
二叉搜索树是一种特殊的二叉树，具有以下性质：
**1. 如果左子树不为空，左子树上结点的值均小于根节点的值
2. 如果右子树不为空，右子树结点上的值均大于根节点的值
3. 左右子树也分别为二叉搜索树**

通过性质我们得知，左子树<根节点<右子树。我们学过三种树的遍历，中序遍历，则刚好是从左子树到根节点到右子树遍历，即可得到一个递增的序列，然后找到第k小的那个元素即可。

递归方法:

/**
 * 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 result = 0;
        int index = 0;
        if(k < 1)
          return -1;
        InOrder(root, k, index, result);
        return result;
    }

    void InOrder(TreeNode *root, int k, int &index, int &result) {
        if (root) {
            InOrder(root->left, k, index, result);
            index++;//计数
            if (index == k) {
                result = root->val;
            }
            InOrder(root->right, k, index, result);
        }
    }
};

非递归方法：

class Solution {    //非递归方法：
public:
    int kthSmallest(TreeNode *root, int k) {
        int index = 0;
        TreeNode *cur = root;
        stack<TreeNode*> s;
        while (cur || !s.empty()){
            while (cur){
                s.push(cur);
                cur = cur->left;
            }
            cur = s.top();  //返回栈顶元素
            s.pop();
            index++;
            if (index == k)
                return cur->val;
            cur = cur->right;
        }
        return 0;
    }
};