BST Iterator Design in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the stack in the BST iterator constructor.

DSA Typescript

class BSTIterator {
  stack: TreeNode[] = [];
  constructor(root: TreeNode | null) {
    this._leftmostInorder(root);
  }

  private _leftmostInorder(node: TreeNode | null) {
    while (node !== null) {
      this.stack.[1](node);
      node = node.left;
    }
  }
}

Drag options to blanks, or click blank then click option'

Apush

Bpop

Cshift

Dunshift

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop instead of push

Using shift or unshift which are queue operations

We use 'push' to add nodes to the stack as we traverse left.

2fill in blank

medium

Complete the code to check if the iterator has a next smallest number.

DSA Typescript

hasNext(): boolean {
  return this.stack.[1] > 0;
}

Drag options to blanks, or click blank then click option'

Asize

Bcount

Clength

Dcapacity

Attempts:

3 left

💡 Hint

Common Mistakes

Using size which is for sets or maps

Using count or capacity which are not valid array properties

The length property gives the number of elements in the array (stack).

3fill in blank

hard

Fix the error in the next() method to correctly return the next smallest number.

DSA Typescript

next(): number {
  const topmostNodeVal = this.stack.[1]();
  return topmostNodeVal!.val;
}

Drag options to blanks, or click blank then click option'

Apush

Bpop

Cshift

Dunshift

Attempts:

3 left

💡 Hint

Common Mistakes

Using push which adds instead of removes

Using shift or unshift which operate on the start of the array

We use pop to remove and return the top element from the stack.

4fill in blank

hard

Fill both blanks to add the right child nodes to the stack after popping the top node.

DSA Typescript

next(): number {
  const topmostNodeVal = this.stack.pop();
  if (topmostNodeVal !== undefined) {
    let node = topmostNodeVal.[1];
    while (node !== null) {
      this.stack.[2](node);
      node = node.left;
    }
  }
  return topmostNodeVal!.val;
}

Drag options to blanks, or click blank then click option'

Aright

Bleft

Cpush

Dpop

Attempts:

3 left

💡 Hint

Common Mistakes

Using left instead of right for the child

Using pop instead of push to add nodes

After popping, we move to the right child and push its left descendants.

5fill in blank

hard

Fill all three blanks to complete the BST iterator class with constructor, hasNext, and next methods.

DSA Typescript

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

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

  private _leftmostInorder(node: TreeNode | null) {
    while (node !== null) {
      this.stack.[1](node);
      node = node.left;
    }
  }

  hasNext(): boolean {
    return this.stack.[2] > 0;
  }

  next(): number {
    const topmostNode = this.stack.[3]();
    if (topmostNode.right !== null) {
      this._leftmostInorder(topmostNode.right);
    }
    return topmostNode.val;
  }
}

Drag options to blanks, or click blank then click option'

Apush

Bpop

Clength

Dsize

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop instead of push in _leftmostInorder

Using size instead of length for array

Using push instead of pop in next

Push nodes in _leftmostInorder, check stack length in hasNext, pop nodes in next.

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