Introduction

In this blog post, we will explore how to solve the LeetCode problem "173" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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?

Explanation

• In-order Traversal with Stack: Use a stack to simulate the in-order traversal. The stack will store the nodes that are on the path from the root to the current node in the traversal.
  • Lazy Evaluation: Only push nodes onto the stack as needed when next() is called. This ensures O(h) space complexity and amortized O(1) time complexity for next() and hasNext().
  • Initialization: In the constructor, push all the leftmost nodes from the root onto the stack.

• Runtime Complexity: Amortized O(1) for next() and hasNext().
• Storage Complexity: O(h), where h is the height of the BST.

Code

    class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

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

    def _push_left(self, node: TreeNode):
        while node:
            self.stack.append(node)
            node = node.left

    def next(self) -> int:
        node = self.stack.pop()
        self._push_left(node.right)
        return node.val

    def hasNext(self) -> bool:
        return len(self.stack) > 0