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();
*/

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