BST Iterator Design in DSA Java - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main purpose of a BST Iterator?

A BST Iterator allows traversing a Binary Search Tree (BST) in sorted order (in-order traversal) one element at a time, without storing all elements at once.

intermediate

How does a BST Iterator typically achieve in-order traversal without recursion?

It uses a stack to keep track of nodes. It pushes left children down to the stack, then visits nodes and moves to right children, simulating recursion iteratively.

beginner

In BST Iterator design, what does the 'next()' method do?

The 'next()' method returns the next smallest element in the BST and advances the iterator to the following element.

intermediate

What is the time complexity of 'next()' and 'hasNext()' methods in a well-designed BST Iterator?

'next()' and 'hasNext()' both run in average O(1) time, with occasional O(h) time where h is the tree height, due to stack operations.

intermediate

Why is a stack used in BST Iterator design instead of storing all elements in a list?

Using a stack saves memory by only storing nodes needed for the current traversal path, avoiding the need to store the entire BST elements upfront.

What data structure is commonly used inside a BST Iterator to simulate in-order traversal?

A Array

B Queue

C Stack

D HashMap

A stack is used to keep track of nodes during in-order traversal without recursion.

What does the 'hasNext()' method in a BST Iterator check?

A If the current node has a right child

B If the BST is empty

C If the stack is empty

D If there is a next smallest element available

'hasNext()' returns true if there is a next smallest element to return.

What is the order of elements returned by a BST Iterator?

A Pre-order

B In-order

C Post-order

D Level-order

BST Iterator returns elements in sorted order using in-order traversal.

Which of these is true about the space complexity of a BST Iterator using a stack?

A O(h) where h is tree height

B O(n) where n is total nodes

C O(1) constant space

D O(log n) always

The stack stores nodes along the path, so space is proportional to tree height h.

When 'next()' is called, what does the BST Iterator do after returning the current node's value?

A Pushes all left children of the right subtree onto the stack

B Clears the stack

C Moves to the root node

D Returns null

After returning a node, it pushes all left children of that node's right subtree to the stack.

Explain how a BST Iterator uses a stack to perform in-order traversal step-by-step.

Think about how recursion is simulated using a stack.

Describe the advantages of using a BST Iterator over converting the BST to a sorted list first.

Consider memory and performance benefits.

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