BST Iterator Design in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA Java

public BSTIterator(TreeNode root) {
    stack = new Stack<>();
    [1](root);
}

private void pushLeft(TreeNode node) {
    while (node != null) {
        stack.push(node);
        node = node.left;
    }
}

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

ApopRight

BpopLeft

CpushRight

DpushLeft

Attempts:

3 left

💡 Hint

Common Mistakes

Using popLeft instead of pushLeft causes runtime errors.

Calling pushRight does not initialize the stack correctly.

The constructor calls pushLeft(root) to push all left children onto the stack, preparing for in-order traversal.

2fill in blank

medium

Complete the code to check if the BST Iterator has a next element.

DSA Java

public boolean hasNext() {
    return [1].isEmpty() == false;
}

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

Astack

Barray

Clist

Dqueue

Attempts:

3 left

💡 Hint

Common Mistakes

Checking a queue or list instead of the stack.

Using isEmpty() without negation.

The stack stores nodes to visit; if it's not empty, there is a next element.

3fill in blank

hard

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

DSA Java

public int next() {
    TreeNode node = stack.[1]();
    pushLeft(node.right);
    return node.val;
}

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

Apush

Bpeek

Cpop

Dremove

Attempts:

3 left

💡 Hint

Common Mistakes

Using peek() returns the node but does not remove it, causing infinite loops.

Using push() or remove() causes errors.

pop() removes and returns the top node from the stack, which is the next smallest element.

4fill in blank

hard

Fill both blanks to complete the pushLeft method that pushes all left children onto the stack.

DSA Java

private void pushLeft(TreeNode [1]) {
    while ([2] != null) {
        stack.push([2]);
        [2] = [2].left;
    }
}

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

Anode

Broot

Ccurrent

Dtemp

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names causing undefined variable errors.

Using 'root' or 'current' inconsistently.

The method uses 'node' as the parameter and iterates using the same variable to push left children.

5fill in blank

hard

Fill all three blanks to complete the next() method that returns the next smallest element and updates the stack.

DSA Java

public int next() {
    TreeNode [1] = stack.[2]();
    [3]([1].right);
    return [1].val;
}

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

Anode

Bpop

CpushLeft

Dpeek

Attempts:

3 left

💡 Hint

Common Mistakes

Using peek() instead of pop() causes infinite loops.

Not pushing left children of the right subtree.

The method pops the top node as 'node', pushes its right child's left descendants, then returns its value.

Practice

1. What is the main purpose of a BST Iterator in the context of a Binary Search Tree?

easy

A. To traverse the BST in ascending order without recursion

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 using recursion explicitly.

2. Step 2: Identify the main purpose

   It provides a way to get the next smallest element efficiently, which matches To traverse the BST in ascending order without recursion.

3. Final Answer:

   To traverse the BST in ascending order without recursion -> Option A

4. Quick Check:

   BST Iterator = Ascending traversal [OK]

Hint: BST Iterator returns next smallest element in sorted order [OK]

Common Mistakes:

• Confusing iterator with insertion or deletion operations
• Thinking it balances the tree automatically
• Assuming it uses recursion internally

2. Which of the following is the correct method signature for the next() method in a BST Iterator class in Java?

easy

A. public Node next()

B. public void next()

C. public int next()

D. public boolean next()

Solution

1. Step 1: Understand the return type of next()

   The next() method returns the next smallest integer value from the BST, so it should return an int.

2. Step 2: Match method signature

   public int next() matches the expected signature: public int next().

3. Final Answer:

   public int next() -> Option C

4. Quick Check:

   next() returns int value [OK]

Hint: next() returns next smallest int value from BST [OK]

Common Mistakes:

• Using void return type for next()
• Returning boolean instead of int
• Returning Node instead of int value

3. Given the following BST Iterator code snippet, what will be the output sequence of next() calls?

BSTIterator iterator = new BSTIterator(root); // BST with nodes 3, 7, 9
System.out.print(iterator.next() + ", ");
System.out.print(iterator.next() + ", ");
System.out.print(iterator.next());

medium

A. 7, 3, 9

B. 3, 7, 9

C. 9, 7, 3

D. 3, 9, 7

Solution

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

   BST in-order traversal returns nodes in ascending order: 3, 7, 9.

2. Step 2: Trace next() calls

   Each next() call returns the next smallest element, so the output is 3, then 7, then 9.

3. Final Answer:

   3, 7, 9 -> Option B

4. Quick Check:

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

Hint: next() returns ascending BST values in order [OK]

Common Mistakes:

• Assuming next() returns descending order
• Mixing up node values order
• Confusing next() with hasNext()

4. Identify the bug in this BST Iterator next() method implementation:

public int next() {
    TreeNode node = stack.pop();
    if (node.right != null) {
        pushLeft(node.right);
    }
    return node.val;
}

private void pushLeft(TreeNode node) {
    while (node != null) {
        stack.push(node);
        node = node.left;
    }
}

medium

A. The next() method should return node.right.val instead of node.val

B. The pushLeft method should push right children instead of left

C. The stack is not initialized before use

D. The next() method should check if stack is empty before popping

Solution

1. Step 1: Analyze next() method stack usage

   Calling stack.pop() without checking if stack is empty can cause runtime error if no elements remain.

2. Step 2: Identify missing safety check

   Proper implementation should check stack emptiness before popping to avoid exceptions.

3. Final Answer:

   The next() method should check if stack is empty before popping -> Option D

4. Quick Check:

   Check stack empty before pop [OK]

Hint: Always check stack not empty before pop in next() [OK]

Common Mistakes:

• Ignoring stack empty check causing exceptions
• Misunderstanding pushLeft logic
• Returning wrong node value

5. You want to design a BST Iterator that supports peek() to see the next smallest element without advancing the iterator. Which approach correctly extends the standard BST Iterator design?

hard

A. Store the next smallest element in a variable during next() and return it in peek() without popping stack

B. Use an extra stack to store all nodes and pop from it in peek()

C. Modify next() to return the top of the stack without popping it, and peek() pops the stack

D. Traverse the entire BST in constructor and store values in a list for peek() and next()

Solution

1. Step 1: Understand peek() requirement

   peek() should return next smallest element without advancing iterator, so it must not pop stack.

2. Step 2: Identify correct approach

   Storing next smallest element in a variable during next() allows peek() to return it without changing stack state, matching Store the next smallest element in a variable during next() and return it in peek() without popping stack.

3. Final Answer:

   Store the next smallest element in a variable during next() and return it in peek() without popping stack -> Option A

4. Quick Check:

   peek() returns next without advancing [OK]

Hint: Cache next element to support peek() without stack pop [OK]

Common Mistakes:

• Popping stack in peek() advancing iterator
• Using extra stack unnecessarily
• Precomputing all values losing iterator benefits