Why BST Iterator Design in DSA Typescript?

The Big Idea

What if you could explore a huge tree one step at a time without ever losing your place?

The Scenario

Imagine you have a huge family tree drawn on paper, and you want to look at each member one by one in order from youngest to oldest. Doing this manually means flipping through pages and trying to remember where you left off.

The Problem

Manually tracking your place in a big tree is slow and confusing. You might lose track, repeat members, or miss some. It's hard to remember where you stopped without a clear system.

The Solution

A BST Iterator acts like a bookmark that remembers exactly where you are in the tree. It lets you move step-by-step in order without losing your place, making the process smooth and error-free.

Before vs After

✗ Before

function inorderTraversal(root) {
  if (!root) return;
  inorderTraversal(root.left);
  console.log(root.val);
  inorderTraversal(root.right);
}
// No way to pause and resume easily

✓ After

class BSTIterator {
  constructor(root) { /* initialize stack */ }
  next() { /* return next smallest */ }
  hasNext() { /* check if more nodes */ }
}
// Can pause and resume traversal anytime

What It Enables

It enables smooth, step-by-step access to tree elements in order without redoing work or losing track.

Real Life Example

Like reading a book with a bookmark, a BST Iterator helps you explore a large sorted list of contacts one by one without starting over each time.

Key Takeaways

Manual tree traversal is hard to pause and resume.

BST Iterator remembers your place and moves stepwise.

This makes ordered access efficient and easy.

Practice

1. What is the main purpose of a BST Iterator in a binary search tree?

easy

A. To traverse the BST in ascending order without storing all elements at once

B. To insert new nodes into the BST

C. To delete nodes from the BST

D. To balance the BST automatically

Solution

1. Step 1: Understand BST Iterator functionality

   A BST Iterator is designed to traverse the tree in sorted order, typically ascending, without loading all nodes into memory at once.

2. Step 2: Compare options with BST Iterator purpose

   Options B, C, and D relate to modifying the tree, not iterating over it. To traverse the BST in ascending order without storing all elements at once correctly describes the iterator's purpose.

3. Final Answer:

   To traverse the BST in ascending order without storing all elements at once -> Option A

4. Quick Check:

   BST Iterator = In-order traversal on demand [OK]

Hint: BST Iterator returns next smallest element step-by-step [OK]

Common Mistakes:

• Confusing iterator with insertion or deletion operations
• Thinking iterator balances the tree
• Assuming iterator stores all nodes at once

2. Which of the following is the correct TypeScript method signature for the next() method in a BST Iterator class that returns the next smallest number?

easy

A. next(value: number): number

B. next(): void

C. next(): number

D. next(): boolean

Solution

1. Step 1: Understand the purpose of next()

   The next() method should return the next smallest number in the BST, so it must return a number.

2. Step 2: Analyze method signatures

   next(): number returns a number with no parameters, which fits the iterator pattern. next(): void returns void, so no value is returned. next(value: number): number takes a parameter, which is not needed. next(): boolean returns boolean, which is incorrect.

3. Final Answer:

   next(): number -> Option C

4. Quick Check:

   next() returns next smallest number [OK]

Hint: next() returns a number without parameters [OK]

Common Mistakes:

• Adding unnecessary parameters to next()
• Returning void instead of number
• Returning boolean instead of number

3. Given the following BST Iterator usage in TypeScript, what will be the output printed?

const root = {
  val: 7,
  left: { val: 3, left: null, right: null },
  right: { val: 15, left: { val: 9, left: null, right: null }, right: { val: 20, left: null, right: null } }
};

const iterator = new BSTIterator(root);
console.log(iterator.next());
console.log(iterator.next());
console.log(iterator.hasNext());
console.log(iterator.next());

medium

A. [3, 7, true, 9]

B. [7, 3, true, 9]

C. [3, 7, false, 9]

D. [3, 9, true, 15]

Solution

1. Step 1: Understand BST in-order traversal order

   The BST nodes in ascending order are 3, 7, 9, 15, 20.

2. Step 2: Trace iterator calls

   First next() returns 3, second next() returns 7, hasNext() returns true because more nodes exist, third next() returns 9.

3. Final Answer:

   [3, 7, true, 9] -> Option A

4. Quick Check:

   In-order traversal = 3,7,9,... [OK]

Hint: Remember BST in-order is sorted ascending [OK]

Common Mistakes:

• Mixing up left and right subtree order
• Assuming hasNext() returns next value
• Confusing node values order

4. Consider this partial TypeScript BST Iterator class code. What is the bug?

class BSTIterator {
  stack: TreeNode[] = [];

  constructor(root: TreeNode | null) {
    this.pushLeft(root);
  }

  pushLeft(node: TreeNode | null) {
    while (node !== null) {
      this.stack.push(node);
      node = node.right;  // Bug here
    }
  }

  next(): number {
    const node = this.stack.pop()!;
    this.pushLeft(node.right);
    return node.val;
  }

  hasNext(): boolean {
    return this.stack.length > 0;
  }
}

medium

A. next() should pop from the front, not the end

B. In pushLeft, should traverse left child, not right

C. hasNext() should check if stack is empty incorrectly

D. Constructor should not call pushLeft

Solution

1. Step 1: Analyze pushLeft method

   pushLeft should push all left children to stack to simulate in-order traversal. The code incorrectly moves to node.right instead of node.left.

2. Step 2: Check other methods

   next() correctly pops from stack and pushes left children of right subtree. hasNext() correctly checks stack length. Constructor correctly calls pushLeft.

3. Final Answer:

   In pushLeft, should traverse left child, not right -> Option B

4. Quick Check:

   pushLeft must go left to simulate in-order [OK]

Hint: pushLeft must follow left children, not right [OK]

Common Mistakes:

• Traversing right child in pushLeft instead of left
• Misunderstanding stack usage in iterator
• Incorrect pop direction in next()

5. You want to design a BST Iterator that supports peek() to see the next smallest element without advancing the iterator. Which approach best supports this feature efficiently?

hard

A. Use a queue to store nodes in BFS order

B. Store all BST elements in an array and use an index pointer

C. Traverse the BST every time peek() is called

D. Use a stack to store left children and keep the top as peek value

Solution

1. Step 1: Understand peek() requirement

   peek() must return next smallest element without moving forward, so the next element must be accessible without popping.

2. Step 2: Evaluate approaches

   Use a stack to store left children and keep the top as peek value uses a stack with left children; the top of the stack is the next smallest element, so peek() can return it efficiently. Store all BST elements in an array and use an index pointer stores all elements but uses extra memory. Traverse the BST every time peek() is called is inefficient as it traverses repeatedly. Use a queue to store nodes in BFS order uses BFS order, which is not sorted.

3. Final Answer:

   Use a stack to store left children and keep the top as peek value -> Option D

4. Quick Check:

   Stack top = next smallest for peek() [OK]

Hint: Stack top always holds next smallest for peek() [OK]

Common Mistakes:

• Using BFS queue which is unordered
• Re-traversing tree on every peek() call
• Storing all elements wastes memory