# 173. Binary Search Tree Iterator

Difficulty: Medium

Frequency: N/A

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.

My solution:
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Code:
```/**
* 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*> myStack;
void pushLeft(TreeNode *node) {
for (; node != NULL; node = node->left) myStack.push(node);
}
public:
BSTIterator(TreeNode *root) {
pushLeft(root);
}

/** @return whether we have a next smallest number */
bool hasNext() {
return !myStack.empty();
}

/** @return the next smallest number */
int next() {
TreeNode *tmp = myStack.top();
myStack.pop();
pushLeft(tmp->right);
return tmp->val;
}
};

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

Another solution:
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