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.

/**

 * Definition for binary tree

 * struct TreeNode {

 *     int val;

 *     TreeNode *left;

 *     TreeNode *right;

 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}

 * };

 */

class BSTIterator {

public:

    BSTIterator(TreeNode *root) {

        i=0;

		if(NULL==root)

		{

			length=0;

		}

		else

		{

			stack<TreeNode*> nodes;

			TreeNode* cur=root;

			while(!nodes.empty() || cur)

			{

				while(cur)

				{

					nodes.push(cur);

					cur=cur->left;

				}

				cur=nodes.top();

				nodesValue.push_back(cur->val);

				nodes.pop();

				cur=cur->right;

			}

			length=nodesValue.size();

		}

    }

    /** @return whether we have a next smallest number */

    bool hasNext() {

        if(i<length)

			return true;

		return false;

    }

    /** @return the next smallest number */

    int next() {

        return nodesValue[i++];

    }

private:

	vector<int>  nodesValue;

	int i;

	int length;

};

/**

 * Your BSTIterator will be called like this:

 * BSTIterator i = BSTIterator(root);

 * while (i.hasNext()) cout << i.next();

 */