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();
 */