0173__binary_search_tree_iterator.py

"""
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, 10^5].
* 0 <= Node.val <= 10^6
* At most 10^5 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?
"""
from typing import Optional
from unittest import TestCase

from lib.TreeNode import TreeNode, build_tree


class BSTIterator():
    def __init__(self, root: Optional[TreeNode]):
        self.cur = root
        self.stack = []

    def next(self) -> int:
        while self.cur:
            self.stack.append(self.cur)
            self.cur = self.cur.left

        self.cur = self.stack.pop()
        result = self.cur.val
        self.cur = self.cur.right

        return result

    def hasNext(self) -> bool:
        return self.cur is not None or len(self.stack) > 0


class TestIterator(TestCase):
    def test_it(self):
        iterator = BSTIterator(build_tree([7, 3, 15, None, None, 9, 20]))
        self.assertEqual(3, iterator.next())
        self.assertEqual(7, iterator.next())
        self.assertTrue(iterator.hasNext())
        self.assertEqual(9, iterator.next())
        self.assertTrue(iterator.hasNext())
        self.assertEqual(15, iterator.next())
        self.assertTrue(iterator.hasNext())
        self.assertEqual(20, iterator.next())
        self.assertFalse(iterator.hasNext())