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
  

  挑战

  额外空间复杂度是O(h)，其中h是这棵树的高度

  Super Star：使用O(1)的额外空间复杂度

  /**
   * 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:
   * Solution iterator = Solution(root);
   * while (iterator.hasNext()) {
   *    TreeNode * node = iterator.next();
   *    do something for node
   */
  class Solution {
  public:
      //@param root: The root of binary tree.
      Solution(TreeNode *root) {
          // write your code here
          TreeNode *cur = root;
  		while (cur != NULL)
  		{
  			buf.push(cur);
  			cur = cur->left;
  		}
      }
  
      //@return: True if there has next node, or false
      bool hasNext() {
          // write your code here
          return buf.empty()? false : true;
      }
      
      //@return: return next node
      TreeNode* next() {
          // write your code here
          if (buf.empty())
  		{
  			return NULL;
  		}
  		TreeNode *top = buf.top();
  		buf.pop();
  		if (top->right)
  		{
  			TreeNode *cur = top->right;
  			buf.push(cur);
  			cur = cur->left;
  			while (cur)
  			{
  				buf.push(cur);
  				cur = cur->left;
  			}
  		}
  
  		return top;
      }
  private:
  	stack<TreeNode*> buf;
  };