设计实现一个带有下列属性的二叉查找树的迭代器：
元素按照递增的顺序被访问（比如中序遍历）
next()和hasNext()的询问操作要求均摊时间复杂度是O(1)
样例
对于下列二叉查找树，使用迭代器进行中序遍历的结果为 [1, 6, 10, 11, 12]

  10
 /    \
1      11
 \       \
  6       12

题目链接：http://www.lintcode.com/zh-cn/problem/binary-search-tree-iterator/
直接上代码：

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 * Example of iterate a tree:
 * BSTIterator iterator = BSTIterator(root);
 * while (iterator.hasNext()) {
 *    TreeNode * node = iterator.next();
 *    do something for node
 */
class BSTIterator {
public:
    stack<TreeNode *> s;
    //@param root: The root of binary tree.
    BSTIterator(TreeNode *root) {
        // write your code here
        while (root) {
            s.push(root);
            root = root->left;
        }
    }

    //@return: True if there has next node, or false
    bool hasNext() {
        // write your code here
        return !s.empty();
    }
    
    //@return: return next node
    TreeNode* next() {
        // write your code here
        TreeNode *n = s.top();
        s.pop();
        TreeNode* res = n;
        if (n->right) {
            n = n->right;
            while(n) {
                s.push(n);
                n = n->left;
            }
        }
        return res;
    }
};