BST Iterator Design in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare the stack used in the BST Iterator class.

DSA C++

std::stack<TreeNode*> [1];

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

Anodes

Bstack

Celements

Ddata

Attempts:

3 left

💡 Hint

Common Mistakes

Using a generic name like 'data' which is less descriptive.

Using 'stack' as variable name which can be confusing with the type.

The stack is commonly named 'nodes' to hold TreeNode pointers during iteration.

2fill in blank

medium

Complete the code to push all left children of a node onto the stack.

DSA C++

void pushLeft(TreeNode* node) {
    while (node != nullptr) {
        [1].push(node);
        node = node->left;
    }
}

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Anodes

Bstack

Celements

Ddata

Attempts:

3 left

💡 Hint

Common Mistakes

Using the type name 'stack' instead of the variable name.

Using an undefined variable name.

The stack named 'nodes' is used to push nodes while traversing left children.

3fill in blank

hard

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

DSA C++

int next() {
    TreeNode* node = [1].top();
    [1].pop();
    pushLeft(node->right);
    return node->val;
}

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Aelements

Bnodes

Cstack

Ddata

Attempts:

3 left

💡 Hint

Common Mistakes

Using the type name 'stack' instead of the variable name.

Forgetting to pop the node after accessing top.

The stack variable 'nodes' holds the nodes to visit; use it to get the top node.

4fill in blank

hard

Fill both blanks to complete the hasNext() method that checks if more nodes exist.

DSA C++

bool hasNext() {
    return ![1].[2]();
}

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Anodes

Bempty

Csize

Dtop

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'size()' instead of 'empty()' which returns a number, not boolean.

Using 'top()' which returns the top element, not a boolean.

We check if the stack 'nodes' is not empty to know if more nodes exist.

5fill in blank

hard

Fill all three blanks to complete the BST Iterator constructor initializing the stack.

DSA C++

BSTIterator(TreeNode* root) {
    [1] = std::stack<TreeNode*>();
    [2](root);
    }

private:
void [3](TreeNode* node);

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

Anodes

BpushLeft

Dstack

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'stack' as variable name which conflicts with type.

Using different names for the pushLeft method.

The constructor initializes the stack 'nodes' and calls 'pushLeft' to push left nodes. The private method is named 'pushLeft'.

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