BST Iterator Design in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the stack in the BST iterator constructor.

DSA C

void bstIteratorInit(BSTIterator* iter, TreeNode* root) {
    iter->stack = createStack();
    TreeNode* current = root;
    while (current != NULL) {
        stackPush(iter->stack, [1]);
        current = current->left;
    }
}

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

Acurrent->left

Broot

Ccurrent

Diter

Attempts:

3 left

💡 Hint

Common Mistakes

Pushing current->left instead of current causes skipping nodes.

Pushing root repeatedly instead of current causes infinite loop.

We push the current node itself onto the stack to traverse the leftmost path.

2fill in blank

medium

Complete the code to check if the BST iterator has a next smallest element.

DSA C

bool bstIteratorHasNext(BSTIterator* iter) {
    return [1];
}

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Aiter->stack == NULL

BstackIsEmpty(iter->stack)

Citer == NULL

DstackIsEmpty(iter->stack) == false

Attempts:

3 left

💡 Hint

Common Mistakes

Returning true when stack is empty causes wrong hasNext result.

Checking iter == NULL instead of stack emptiness.

If the stack is not empty, there is a next smallest element.

3fill in blank

hard

Fix the error in the bstIteratorNext function to correctly return the next smallest value.

DSA C

int bstIteratorNext(BSTIterator* iter) {
    TreeNode* node = stackPop(iter->stack);
    int result = node->val;
    node = node->right;
    while (node != NULL) {
        stackPush(iter->stack, [1]);
        node = node->left;
    }
    return result;
}

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Anode

Bnode->right

Citer->stack

Dresult

Attempts:

3 left

💡 Hint

Common Mistakes

Pushing node->right causes infinite loop or wrong traversal.

Pushing result (int) instead of node pointer causes type errors.

We push the current node pointer to traverse its left subtree.

4fill in blank

hard

Fill both blanks to complete the stack push and pop operations in the BST iterator.

DSA C

void stackPush(Stack* stack, [1] value) {
    Node* newNode = malloc(sizeof(Node));
    newNode->value = value;
    newNode->next = stack->top;
    stack->top = newNode;
}

[2] stackPop(Stack* stack) {
    if (stack->top == NULL) return NULL;
    Node* temp = stack->top;
    [3] value = temp->value;
    stack->top = temp->next;
    free(temp);
    return value;
}

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ATreeNode*

Bint

Dvoid

Attempts:

3 left

💡 Hint

Common Mistakes

Using int type causes type mismatch errors.

Using void as return type for pop causes missing return value.

The stack stores TreeNode pointers, so push and pop use TreeNode* type.

5fill in blank

hard

Fill all three blanks to complete the BST iterator next function with correct stack operations and return value.

DSA C

int bstIteratorNext(BSTIterator* iter) {
    TreeNode* node = [1](iter->stack);
    int val = node->val;
    node = node->right;
    while (node != NULL) {
        [2](iter->stack, node);
        node = node->left;
    }
    return [3];
}

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

AstackPop

BstackPush

Cval

DstackTop

Attempts:

3 left

💡 Hint

Common Mistakes

Using stackTop instead of stackPop causes infinite loop.

Returning node instead of val causes wrong return type.

Pop the top node, push left children of right subtree, and return the node's value.

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