BST Iterator Design in DSA C - Time & Space Complexity

Time Complexity: BST Iterator Design

O(1) amortized per next()

Understanding Time Complexity

We want to understand how fast the BST Iterator works when we move through the tree.

How does the time to get the next smallest item change as the tree grows?

Scenario Under Consideration

Analyze the time complexity of the following BST Iterator code snippet.

// BST Iterator structure
struct BSTIterator {
    Stack* stack;
};

// Initialize iterator with root
void bstIteratorInit(BSTIterator* it, TreeNode* root) {
    while (root != NULL) {
        stackPush(it->stack, root);
        root = root->left;
    }
}

// Get next smallest element
int bstIteratorNext(BSTIterator* it) {
    TreeNode* node = stackPop(it->stack);
    int val = node->val;
    node = node->right;
    while (node != NULL) {
        stackPush(it->stack, node);
        node = node->left;
    }
    return val;
}
    

This code initializes the iterator by pushing all left children, then each next call pops the top node and pushes left children of its right subtree.

Identify Repeating Operations

Look at what repeats when we call next.

• Primary operation: Pushing nodes onto the stack while going left.
• How many times: Each node is pushed and popped exactly once during the entire iteration.

How Execution Grows With Input

Each node is handled once, but some next calls do more work pushing nodes.

Input Size (n)	Approx. Operations
10	About 10 pushes and 10 pops spread over calls
100	About 100 pushes and 100 pops total
1000	About 1000 pushes and 1000 pops total

Pattern observation: Total work grows linearly with number of nodes, but each next call may vary in cost.

Final Time Complexity

Time Complexity: O(1) amortized per next()

This means each next call takes constant time on average, even if some calls do more work.

Common Mistake

[X] Wrong: "Each next() call always takes O(h) time where h is tree height."

[OK] Correct: Some calls do O(h) work, but over all nodes, the total work spreads out, making average next() call O(1).

Interview Connect

Understanding amortized time helps you explain why this iterator is efficient and practical for large trees.

Self-Check

What if we changed the iterator to push all nodes at once during initialization? How would the time complexity of next() change?

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