Binary Search Tree Iterator

Implement an iterator over a binary search tree (BST). Your iterator will be initialized with the root node of a BST.

Calling next() will return the next smallest number in the BST.

Note: next() and hasNext() should run in average O(1) time and uses O(h) memory, where h is the height of the tree.

采用中序遍历将节点压入栈中，由于要求存储空间为O（h），因此不能在一开始将所有节点全部压入，只是压入最左边一列。当取出一个节点时，压入其右子节点的所有左节点。

 /**
  * Definition for binary tree
  * struct TreeNode {
  *     int val;
  *     TreeNode *left;
  *     TreeNode *right;
  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
  * };
  */
 class BSTIterator {
 private:
     stack <TreeNode*> stk;
     int minres;
 public:
     BSTIterator(TreeNode *root) {
         while(root)
         {
             stk.push(root);
             root=root->left;
         }
     }

     /** @return whether we have a next smallest number */
     bool hasNext() {
         if(stk.empty())
             return false;
         TreeNode* top;
         top=stk.top();
         minres=top->val;
         stk.pop();
         TreeNode* cur=top->right;
         if(cur)
         {
             stk.push(cur);
             cur=cur->left;
             while(cur)
             {
                 stk.push(cur);
                 cur=cur->left;
             }
         }
         return true;
     }

     /** @return the next smallest number */
     int next() {
         return minres;
     }
 };

 /**
  * Your BSTIterator will be called like this:
  * BSTIterator i = BSTIterator(root);
  * while (i.hasNext()) cout << i.next();
  */