Implement the `BSTIterator` class that represents an iterator over the in-order traversal of a binary search tree (BST):

- `BSTIterator(TreeNode root)` Initializes an object of the `BSTIterator` class. The `root` of the BST is given as part of the constructor. The pointer should be initialized to a non-existent number smaller than any element in the BST.
- `boolean hasNext()` Returns `true` if there exists a number in the traversal to the right of the pointer, otherwise returns `false`.
- `int next()` Moves the pointer to the right, then returns the number at the pointer.

Notice that by initializing the pointer to a non-existent smallest number, the first call to `next()` will return the smallest element in the BST.

You may assume that `next()` calls will always be valid. That is, there will be at least a next number in the in-order traversal when `next()` is called.

Example 1:

    Input
    ["BSTIterator", "next", "next", "hasNext", "next", "hasNext", "next", "hasNext", "next", "hasNext"]
    [[[7, 3, 15, null, null, 9, 20]], [], [], [], [], [], [], [], [], []]
    Output
    [null, 3, 7, true, 9, true, 15, true, 20, false]

    Explanation
    BSTIterator bSTIterator = new BSTIterator([7, 3, 15, null, null, 9, 20]);
    bSTIterator.next();    // return 3
    bSTIterator.next();    // return 7
    bSTIterator.hasNext(); // return True
    bSTIterator.next();    // return 9
    bSTIterator.hasNext(); // return True
    bSTIterator.next();    // return 15
    bSTIterator.hasNext(); // return True
    bSTIterator.next();    // return 20
    bSTIterator.hasNext(); // return False

Constraints:

- The number of nodes in the tree is in the range `[1, 105]`.
- `0 <= Node.val <= 106`
- At most `105` calls will be made to `hasNext`, and `next`.

Follow up:

- Could you implement `next()` and `hasNext()` to run in average `O(1)` time and use `O(h)` memory, where `h` is the height of the tree?