It is similar to use stack for BST inorder traversal.

So do the same thing : 

1. Get stack to store right branchs (include current node).

2. When you pop the top node, remember to push all the right sub branches again. Otherwise, it will cause right's left branches missed from result

 

1 /** 2  * Definition for binary tree 3  * struct TreeNode { 4  *     int val; 5  *     TreeNode *left; 6  *     TreeNode *right; 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8  * }; 9  */10 class BSTIterator {11 private:12     stack
     
       s;13 public:14     void initialNodes(TreeNode *root) {15         while (root) {16             s.push(root);17             root = root->left;18         }19     }20     BSTIterator(TreeNode *root) {21         initialNodes(root);22     }23 24     /** @return whether we have a next smallest number */25     bool hasNext() {26         return !s.empty();27     }28 29     /** @return the next smallest number */30     int next() {31         TreeNode *tmp = s.top();32         s.pop();33         if (tmp->right) {34             initialNodes(tmp->right);35         }36         return tmp->val;37     }38 };39 40 /**41  * Your BSTIterator will be called like this:42  * BSTIterator i = BSTIterator(root);43  * while (i.hasNext()) cout << i.next();44  */