BST Iterator Design in DSA C - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of BST Iterator next() calls

Given a BST Iterator initialized with the BST below, what is the output sequence of calling next() three times?

DSA C

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

BSTIterator iterator = BSTIterator_create(root);
int a = BSTIterator_next(iterator);
int b = BSTIterator_next(iterator);
int c = BSTIterator_next(iterator);
printf("%d %d %d", a, b, c);

A 3 7 9

B 7 3 9

C 3 9 7

D 9 7 3

💡 Hint

Remember that BST Iterator returns elements in ascending order.

The BST Iterator returns the next smallest element each time next() is called. The in-order traversal of the BST is 3, 7, 9, 15, 20. So the first three next() calls return 3, 7, and 9 respectively.

🧠 Conceptual

intermediate

Space complexity of BST Iterator

What is the worst-case space complexity of a BST Iterator that uses a stack to simulate in-order traversal on a BST with n nodes?

A O(n log n)

B O(log n)

C O(1)

D O(n)

💡 Hint

Consider the shape of the BST and how many nodes can be on the stack at once.

In the worst case, the BST is skewed (like a linked list), so the stack can hold all n nodes. Thus, the space complexity is O(n).

🔧 Debug

advanced

Identify the error in BST Iterator next() implementation

What error will this BST Iterator next() method cause when called on a non-empty BST?

DSA C

int BSTIterator_next(BSTIterator* iter) {
    TreeNode* node = stack_pop(iter->stack);
    if (node->right != NULL) {
        stack_push(iter->stack, node->right);
    }
    return node->val;
}

A It misses pushing all left children of the right subtree, causing incorrect order.

B It returns nodes in descending order, not ascending.

C It causes a segmentation fault due to null pointer dereference.

D It causes infinite loop due to not popping nodes.

💡 Hint

Think about how in-order traversal visits left children before the node and right subtree.

The code pushes only the right child but does not push all left descendants of the right child. This breaks the in-order traversal order, causing incorrect output.

📝 Syntax

advanced

Syntax error in BST Iterator initialization

Which option correctly initializes the stack with all left children of the root node in a BST Iterator constructor?

DSA C

void BSTIterator_init(BSTIterator* iter, TreeNode* root) {
    while (root != NULL) {
        // push root
        root = root->left;
    }
}

A while (root != NULL) { stack_push(iter->stack, root); root = root->right; }

B while (root != NULL) { stack_push(iter->stack); root = root->left; }

C while (root != NULL) { stack_push(iter->stack, root); root = root->left; }

D while (root != NULL) { stack_push(iter->stack, root->val); root = root->left; }

💡 Hint

stack_push needs two arguments: the stack and the node pointer.

Option C correctly pushes the current node pointer onto the stack and moves to the left child. Option C misses the node argument, causing a syntax error. Option C moves right instead of left, breaking logic. Option C pushes node value instead of node pointer, breaking stack usage.

🚀 Application

expert

Number of next() calls to exhaust BST Iterator

Given a BST with 10 nodes, how many times can you call next() before hasNext() returns false?

A Depends on the shape of the BST

B 10

C 11

D 9

💡 Hint

Each next() call returns exactly one node value.

The BST Iterator returns each node exactly once. For 10 nodes, next() can be called 10 times before no nodes remain.

Practice

1. What is the main purpose of a BST Iterator in a binary search tree?

easy

A. To delete nodes from the BST

B. To insert new nodes into the BST

C. To balance the BST automatically

D. To traverse the BST in ascending order without recursion

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.

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 D

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
• Thinking it balances the tree automatically
• Assuming it uses recursion internally

2. Which of the following is the correct way to declare a stack for implementing a BST Iterator in C?

easy

A. int stack[100]; int top = -1;

B. stack<int> s; // C++ syntax

C. List<int> stack = new List<int>();

D. var stack = [];

Solution

1. Step 1: Identify language and syntax

   The question is about C language, so C++ or other language syntax is invalid.

2. Step 2: Choose correct C stack declaration

   int stack[100]; int top = -1; declares a fixed-size integer array and a top index, which is the typical way to implement a stack in C.

3. Final Answer:

   int stack[100]; int top = -1; -> Option A

4. Quick Check:

   C stack = array + top index [OK]

Hint: Use array and top index for stack in C [OK]

Common Mistakes:

• Using C++ or other language syntax in C
• Using dynamic lists not available in C
• Using var keyword which is not in C

3. Consider the following BST Iterator code snippet in C. What will be the output sequence of next() calls?

typedef struct Node {
  int val;
  struct Node* left;
  struct Node* right;
} Node;

// Assume stack and top are global
void push(Node* node) { stack[++top] = node; }
Node* pop() { return stack[top--]; }

void pushLeft(Node* root) {
  while (root) {
    push(root);
    root = root->left;
  }
}

Node* next() {
  Node* node = pop();
  pushLeft(node->right);
  return node;
}

// BST structure:
//       5
//      / \
//     3   7
//    / \   \
//   2   4   8

// Initial call: pushLeft(root) where root = 5

// Calls: next(), next(), next(), next(), next(), next()

medium

A. [2, 4, 3, 5, 8, 7]

B. [5, 3, 2, 4, 7, 8]

C. [2, 3, 4, 5, 7, 8]

D. [8, 7, 5, 4, 3, 2]

Solution

1. Step 1: Understand pushLeft and next behavior

   pushLeft pushes all left children starting from root. next pops top node, then pushes left children of right subtree.

2. Step 2: Trace calls step-by-step

   Initial stack after pushLeft(5): [5,3,2] (top is 2). next() pops 2 (no right child), output 2.
   next() pops 3, pushes left of 4 (which is 4), output 3.
   next() pops 4 (no right child), output 4.
   next() pops 5, pushes left of 7 (which is 7), output 5.
   next() pops 7, pushes left of 8 (which is 8), output 7.
   next() pops 8, output 8.

3. Final Answer:

   [2, 3, 4, 5, 7, 8] -> Option C

4. Quick Check:

   In-order traversal = [2,3,4,5,7,8] [OK]

Hint: Trace stack push/pop for in-order sequence [OK]

Common Mistakes:

• Ignoring pushLeft on right child
• Assuming preorder or postorder output
• Mixing up stack order

4. The following code snippet for BST Iterator has a bug. What is the error?

typedef struct Node {
  int val;
  struct Node* left;
  struct Node* right;
} Node;

Node* stack[100];
int top = -1;

void push(Node* node) { stack[++top] = node; }
Node* pop() { return stack[top--]; }

void pushLeft(Node* root) {
  while (root != NULL) {
    push(root);
    root = root->left;
  }
}

Node* next() {
  Node* node = pop();
  pushLeft(node->right);
  return node;
}

int hasNext() {
  return top >= 0;
}

// Usage:
// pushLeft(root);
// while (hasNext()) {
//   Node* n = next();
//   printf("%d ", n->val);
// }

medium

A. pop() returns NULL when stack is empty causing crash

B. Stack overflow if BST is large and pushLeft pushes too many nodes

C. pushLeft does not handle NULL root correctly

D. hasNext() returns wrong boolean value

Solution

1. Step 1: Analyze stack usage in pushLeft

   pushLeft pushes all left nodes onto stack. For a very deep BST, this can exceed stack size 100 causing overflow.

2. Step 2: Check other functions for errors

   pop() assumes stack not empty; hasNext() correctly checks top >= 0; pushLeft handles NULL root by stopping loop.

3. Final Answer:

   Stack overflow if BST is large and pushLeft pushes too many nodes -> Option B

4. Quick Check:

   Fixed stack size risks overflow [OK]

Hint: Check fixed stack size limits for large BSTs [OK]

Common Mistakes:

• Assuming pop() returns NULL safely
• Thinking pushLeft fails on NULL root
• Believing hasNext() logic is incorrect

5. You want to design a BST Iterator that supports next() and hasNext() in O(1) average time and uses O(h) memory, where h is tree height. Which approach best achieves this?

hard

A. Use a stack to store left children, pushing right subtree nodes on next() calls

B. Store all nodes in an array by in-order traversal at initialization

C. Use recursion to traverse nodes on each next() call

D. Use a queue to store nodes level by level

Solution

1. Step 1: Understand time and space constraints

   O(1) average time per next() and O(h) memory means no full traversal storage and no recursion per call.

2. Step 2: Evaluate options

   Use a stack to store left children, pushing right subtree nodes on next() calls uses a stack to store left children only, pushing right subtree nodes lazily, matching constraints.
   Store all nodes in an array by in-order traversal at initialization uses O(n) memory, violating O(h).
   Use recursion to traverse nodes on each next() call uses recursion each call, not O(1) average.
   Use a queue to store nodes level by level uses queue level order, not in-order traversal.

3. Final Answer:

   Use a stack to store left children, pushing right subtree nodes on next() calls -> Option A

4. Quick Check:

   Stack + lazy push = O(1) avg time & O(h) space [OK]

Hint: Stack with lazy right pushes fits O(1) avg time and O(h) space [OK]

Common Mistakes:

• Precomputing all nodes wastes memory
• Using recursion each next() call is slow
• Using queue gives wrong traversal order