Binary Search Tree Iterator Python Solution

lcpy gen -n 173   # by problem number
lcpy gen -s binary_search_tree_iterator   # by problem name

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

from leetcode_py import TreeNode


class BSTIterator:
    # Time: O(1) average per operation
    # Space: O(h)
    def __init__(self, root: TreeNode[int] | None) -> None:
        self._stack: list[TreeNode[int]] = []
        self._push_left(root)

    def _push_left(self, node: TreeNode[int] | None) -> None:
        while node is not None:
            self._stack.append(node)
            node = node.left

    # Time: O(1) average
    # Space: O(1)
    def next(self) -> int:
        node = self._stack.pop()
        self._push_left(node.right)
        return node.val

    # Time: O(1)
    # Space: O(1)
    def has_next(self) -> bool:
        return len(self._stack) > 0