BST Iterator Design in DSA C++ - Build from Scratch

BST Iterator Design

📖 Scenario: Imagine you have a large collection of numbers stored in a special tree called a Binary Search Tree (BST). You want to look at these numbers one by one in order, like flipping through pages in a book. To do this easily, you will build a tool called a BST Iterator that helps you move through the numbers step-by-step.

🎯 Goal: You will create a BST Iterator in C++ that lets you go through the numbers in the tree from smallest to largest. You will build it step-by-step: first creating the tree, then setting up the iterator, then writing the main logic to move through the tree, and finally printing the numbers in order.

📋 What You'll Learn

Create a BST with given nodes

Implement a stack to help with the iterator

Write the iterator methods to get the next smallest number

Print the numbers in ascending order using the iterator

💡 Why This Matters

🌍 Real World

BST Iterators are useful in databases and search engines where you need to access sorted data efficiently without loading everything at once.

💼 Career

Understanding BST iterators helps in roles involving data structure optimization, system design, and coding interviews.

1

Create the BST nodes

Create a struct called TreeNode with an int val, and two pointers left and right. Then create the BST with these exact nodes and connections:
root with value 7,
root's left child with value 3,
root's right child with value 15,
15's left child with value 9,
15's right child with value 20.

DSA C++

// Define TreeNode struct and create the BST nodes
// Your code here

Hint

Start by defining the TreeNode structure with value and pointers. Then create each node and connect them exactly as described.

2

Set up the BST Iterator class with a stack

Create a class called BSTIterator with a private member std::stack called nodeStack. Add a constructor that takes a TreeNode* called root and calls a private method pushLeft(TreeNode* node) to push all left children of root onto nodeStack.

DSA C++

class BSTIterator {
private:
    std::stack nodeStack;
    void pushLeft(TreeNode* node);
public:
    BSTIterator(TreeNode* root);
    // Your code here
};

Hint

Use a stack to store nodes. The constructor should call pushLeft to add all left nodes starting from root.

3

Implement next() and hasNext() methods

Inside the BSTIterator class, implement the method int next() that pops the top node from nodeStack, calls pushLeft on its right child, and returns its value. Also implement bool hasNext() that returns true if nodeStack is not empty, otherwise false.

DSA C++

class BSTIterator {
private:
    std::stack nodeStack;
    void pushLeft(TreeNode* node) {
        while (node != nullptr) {
            nodeStack.push(node);
            node = node->left;
        }
    }
public:
    BSTIterator(TreeNode* root) {
        pushLeft(root);
    }
    int next();
    bool hasNext();
    // Your code here
};

Hint

Use next() to pop the top node, push left nodes of its right child, and return its value. hasNext() checks if stack is empty.

4

Use the BST Iterator to print all values in ascending order

Create a BSTIterator object called iterator with root. Use a while loop with iterator.hasNext() to print each value returned by iterator.next() separated by spaces.

DSA C++

BSTIterator iterator(root);
// Use while loop to print values in ascending order
// Your code here

Hint

Create the iterator and use a loop to print all values in order separated by spaces.

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 = In-order traversal without recursion [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 way to declare a stack to be used in a BST Iterator in C++?

easy

A. stack s;

B. stack<int> s;

C. stack<TreeNode> s;

D. stack s;

Solution

1. Step 1: Identify the data type stored in the stack

   The BST Iterator stores pointers to tree nodes to traverse the tree efficiently, so the stack should hold TreeNode pointers.

2. Step 2: Match the correct syntax

   stack s; uses stack which is correct. stack<int> s; uses int, which is incorrect. stack<TreeNode> s; uses TreeNode by value, which is inefficient. stack s; uses Node*, which is undefined here.

3. Final Answer:

   stack s; -> Option D

4. Quick Check:

   Stack stores pointers to TreeNode [OK]

Hint: Stack must hold pointers to tree nodes, not values [OK]

Common Mistakes:

• Using stack of int instead of TreeNode*
• Using stack of TreeNode by value causing copies
• Using undefined Node* type

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

struct TreeNode {
  int val;
  TreeNode* left;
  TreeNode* right;
  TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
};

class BSTIterator {
  stack s;
  void pushLeft(TreeNode* node) {
    while (node) {
      s.push(node);
      node = node->left;
    }
  }
public:
  BSTIterator(TreeNode* root) { pushLeft(root); }
  int next() {
    TreeNode* node = s.top(); s.pop();
    pushLeft(node->right);
    return node->val;
  }
  bool hasNext() { return !s.empty(); }
};

// BST structure:
//       7
//      / \
//     3   15
//         / \
//        9  20

BSTIterator it(root);
while(it.hasNext()) {
  cout << it.next() << " ";
}

medium

A. 20 15 9 7 3

B. 7 3 15 9 20

C. 3 7 9 15 20

D. 3 9 7 15 20

Solution

1. Step 1: Understand the BST structure and in-order traversal

   The BST is: 7 / \ 3 15 / \ 9 20 In-order traversal visits nodes in ascending order: 3, 7, 9, 15, 20.

2. Step 2: Trace the iterator's next() calls

   Iterator pushes left nodes first, so initial stack: 7, 3 (top). Calls to next() pop 3, then 7, then push left of 7's right (15), then 9, then 15, then 20. Output sequence matches 3 7 9 15 20.

3. Final Answer:

   3 7 9 15 20 -> Option C

4. Quick Check:

   In-order traversal = 3 7 9 15 20 [OK]

Hint: BST Iterator outputs nodes in ascending order (in-order) [OK]

Common Mistakes:

• Confusing pre-order or post-order traversal output
• Not pushing left children correctly
• Misunderstanding stack top element order

4. Identify the bug in the following BST Iterator next() method and select the correct fix:

int next() {
  TreeNode* node = s.top();
  s.pop();
  pushLeft(node->left); // Bug here
  return node->val;
}

medium

A. Remove pushLeft call completely

B. Change pushLeft(node->left) to pushLeft(node->right)

C. Change s.pop() to s.pop(node)

D. Add check if node is nullptr before pushLeft

Solution

1. Step 1: Analyze the next() method logic

   After popping the current node, the iterator should push all left children of the right subtree to continue in-order traversal.

2. Step 2: Identify the incorrect call

   The code incorrectly calls pushLeft(node->left), which revisits the left subtree already processed. It should call pushLeft(node->right) to process the right subtree.

3. Final Answer:

   Change pushLeft(node->left) to pushLeft(node->right) -> Option B

4. Quick Check:

   Push left of right subtree after pop [OK]

Hint: Push left nodes of right child after popping [OK]

Common Mistakes:

• Pushing left subtree again causing infinite loop
• Removing pushLeft call causing incomplete traversal
• Incorrect pop syntax causing compile error

5. You want to design a BST Iterator that supports a method peek() which returns the next smallest element without advancing the iterator. How would you modify the existing BST Iterator design to support peek() efficiently?

hard

A. Store the next smallest element in a variable during next() and return it in peek()

B. Use an additional stack to store peeked elements separately

C. Modify next() to not pop from stack and peek() to pop instead

D. Precompute all elements in sorted order and store in a vector

Solution

1. Step 1: Understand peek() requirement

   peek() should return next smallest element without moving the iterator forward, so it must remember the next element.

2. Step 2: Modify iterator to store next element

   By storing the next smallest element in a variable during next(), peek() can return this stored value without popping or changing the stack.

3. Step 3: Evaluate other options

   Using an extra stack or precomputing all elements wastes space or changes iterator behavior. Modifying next() to not pop breaks standard iteration.

4. Final Answer:

   Store the next smallest element in a variable during next() and return it in peek() -> Option A

5. Quick Check:

   Cache next element for peek() [OK]

Hint: Cache next element on next() call for peek() [OK]

Common Mistakes:

• Precomputing all elements wastes memory
• Changing next() behavior breaks iteration
• Using extra stack complicates design unnecessarily