Binary Search Tree Iterator Python Solution

Tested Python solution for LeetCode 173 with 15 pytest cases. Generate a practice environment with lcpy.

LeetCode 173, Medium. Topics: Stack, Tree, Design, Binary Search Tree, Binary Tree, Iterator. View on LeetCode.

Generate this problem as a practice environment: tested reference solution, 15 parametrized pytest cases, and a playground notebook:

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

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Problem

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.

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Examples

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

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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?

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Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

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

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Complexity

Time	Space
O(1) average per operation	O(h)

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Tags

NeetCode All.