BST Iterator Design in DSA Java - Time & Space Complexity

Time Complexity: BST Iterator Design

O(1) amortized

Understanding Time Complexity

We want to understand how fast the BST Iterator works when we ask for the next smallest element.

How does the time to get the next element change as the tree grows bigger?

Scenario Under Consideration

Analyze the time complexity of the following BST Iterator methods.

public class BSTIterator {
    private Stack stack = new Stack<>();

    public BSTIterator(TreeNode root) {
        pushLeft(root);
    }

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

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

    public boolean hasNext() {
        return !stack.isEmpty();
    }
}
    

This code creates an iterator over a binary search tree that returns elements in ascending order.

Identify Repeating Operations

Look at what repeats when we call next().

• Primary operation: Popping from the stack and pushing left children of the right subtree.
• How many times: Each node is pushed and popped exactly once during the entire iteration.

How Execution Grows With Input

Getting the next element sometimes takes a few steps to push nodes, sometimes just one pop.

Input Size (n)	Approx. Operations per next()
10	Usually 1 to 2 steps
100	Usually 1 to 3 steps
1000	Usually 1 to 4 steps

Pattern observation: Each node is handled once, so average work per next() call stays small even as tree grows.

Final Time Complexity

Time Complexity: O(1) amortized

This means each call to next() takes constant time on average, even if some calls take longer.

Common Mistake

[X] Wrong: "Each next() call always takes time proportional to the height of the tree."

[OK] Correct: Because the pushing of nodes is spread out over all calls, the average time per call is constant, not always the height.

Interview Connect

Understanding this iterator's time complexity shows you can analyze how work spreads over many operations, a key skill in coding interviews.

Self-Check

What if we changed the iterator to push all nodes into a list at the start? How would the time complexity of next() change?

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