Pop Using Linked List Node in DSA C - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does the 'pop' operation do in a linked list?

It removes the first node (head) from the linked list and returns its value, effectively shortening the list by one node.

beginner

In a singly linked list, what pointer must be updated when popping the head node?

The head pointer must be updated to point to the next node after the current head.

intermediate

Why is it important to free the popped node in C after removing it from the linked list?

To release the memory allocated for that node and avoid memory leaks in the program.

beginner

What should the pop function return if the linked list is empty?

It should return a special value (like NULL or an error code) indicating that there is no node to pop.

intermediate

Show the basic steps to pop a node from a singly linked list in C.

1. Check if the list is empty (head is NULL).
2. Save the current head node in a temporary pointer.
3. Update head to point to the next node.
4. Save the data from the popped node.
5. Free the popped node's memory.
6. Return the saved data.

What happens to the head pointer after popping a node from a linked list?

A It points to the next node after the popped node.

B It remains pointing to the popped node.

C It points to NULL always.

D It points to the last node.

After popping, the head pointer must move to the next node to keep the list connected.

Which of these is a necessary step when popping a node in C?

A Free the memory of the popped node.

B Allocate new memory for the popped node.

C Change the data of the popped node.

D Reverse the linked list.

Freeing the popped node's memory prevents memory leaks.

If the linked list is empty, what should the pop function do?

A Create a new node and return it.

B Pop the last node anyway.

C Return NULL or an error indicator.

D Crash the program.

Returning NULL or an error prevents undefined behavior when popping from an empty list.

What data structure is typically used to implement a pop operation in a linked list?

A Binary tree.

B Array with fixed size.

C Hash table.

D Singly linked list with a head pointer.

Pop usually removes the first node from a singly linked list using the head pointer.

Which pointer is saved temporarily during the pop operation?

A The tail node pointer.

B The current head node pointer.

C The next node pointer after head.

D A random node pointer.

The current head node is saved so it can be freed after updating the head pointer.

Explain the steps to pop a node from the head of a singly linked list in C.

Think about what happens to the head pointer and memory.

Why is it important to handle the empty list case when popping from a linked list?

Consider what happens if you try to pop when no nodes exist.

Practice

1. What does the pop operation do in a linked list implemented as a stack?

easy

A. Deletes the entire linked list

B. Adds a new node at the end of the linked list

C. Removes and returns the first node (top) of the linked list

D. Returns the last node without removing it

Solution

1. Step 1: Understand pop operation in stack

   Pop removes the top element from the stack, which is the first node in a linked list stack.

2. Step 2: Apply to linked list

   In a linked list used as a stack, the first node is removed and returned during pop.

3. Final Answer:

   Removes and returns the first node (top) of the linked list -> Option C

4. Quick Check:

   Pop = remove first node [OK]

Hint: Pop removes the first node in linked list stack [OK]

Common Mistakes:

• Thinking pop removes last node
• Confusing pop with push
• Assuming pop deletes entire list

2. Which of the following is the correct syntax to pop a node from a linked list stack in C?

easy

A. node* pop(node** head) { node* temp = *head; *head = (*head)->next; return temp; }

B. node pop(node head) { head = head->next; return head; }

C. void pop(node* head) { head = head->next; }

D. node* pop(node* head) { head = head->next; return head; }

Solution

1. Step 1: Check parameter passing for head pointer

   To modify the head pointer, we pass a pointer to pointer (node** head).

2. Step 2: Verify pop logic

   Store current head in temp, move head to next node, return temp node.

3. Final Answer:

   node* pop(node** head) { node* temp = *head; *head = (*head)->next; return temp; } -> Option A

4. Quick Check:

   Pointer to pointer needed to update head [OK]

Hint: Use node** to update head pointer in pop [OK]

Common Mistakes:

• Passing head by value instead of pointer to pointer
• Returning wrong node or void
• Not updating head pointer correctly

3. What will be the output of the following code snippet?

typedef struct node { int data; struct node* next; } node;

node* pop(node** head) {
  if (*head == NULL) return NULL;
  node* temp = *head;
  *head = (*head)->next;
  return temp;
}

int main() {
  node n3 = {3, NULL};
  node n2 = {2, &n3};
  node n1 = {1, &n2};
  node* head = &n1;
  node* popped = pop(&head);
  printf("%d -> %d -> %d -> NULL", head->data, head->next->data, head->next->next->data);
  return 0;
}

medium

A. 2 -> 3 -> 0 -> NULL

B. 2 -> 3 -> NULL -> NULL

C. 2 -> 3 -> NULL

D. 2 -> 3 -> (segmentation fault)

Solution

1. Step 1: Trace pop operation

   Pop removes node with data 1, head now points to node with data 2.

2. Step 2: Trace printf arguments

   head->data = 2, head->next->data = 3, head->next->next = NULL, accessing ->data causes segmentation fault after printing "2 -> 3 -> " .

3. Final Answer:

   2 -> 3 -> (segmentation fault) -> Option D

4. Quick Check:

   Accessing NULL->data causes crash [OK]

Hint: Accessing next of last node causes crash [OK]

Common Mistakes:

• Assuming NULL prints as 'NULL' string
• Ignoring segmentation fault on NULL dereference
• Misreading linked list after pop

4. Identify the bug in this pop function and select the correct fix:

node* pop(node* head) {
  node* temp = head;
  head = head->next;
  return temp;
}

medium

A. Pass head as node** and update *head inside function

B. Return head instead of temp

C. Set temp->next = NULL before return

D. No bug, function is correct

Solution

1. Step 1: Analyze head pointer update

   Function updates local copy of head, original head outside remains unchanged.

2. Step 2: Fix pointer update

   Pass pointer to pointer (node**) to modify original head pointer.

3. Final Answer:

   Pass head as node** and update *head inside function -> Option A

4. Quick Check:

   Use node** to update head pointer [OK]

Hint: Use node** to update head pointer in pop [OK]

Common Mistakes:

• Updating local copy of head only
• Not passing pointer to pointer
• Returning wrong node

5. Given a linked list stack with nodes containing integers, write the logic to pop all nodes and print their data in order. Which approach is correct?

hard

A. Traverse list without popping and print data

B. Use a loop: while head != NULL, pop node, print data, free node

C. Pop once, print data, then stop

D. Recursively call pop until head is NULL, then print all data after recursion

Solution

1. Step 1: Understand popping all nodes

   Pop removes top node each time, so loop until list is empty.

2. Step 2: Print and free nodes in loop

   Print data of popped node immediately, then free memory to avoid leaks.

3. Final Answer:

   Use a loop: while head != NULL, pop node, print data, free node -> Option B

4. Quick Check:

   Loop pop-print-free until empty [OK]

Hint: Loop pop until head NULL, print and free each node [OK]

Common Mistakes:

• Recursing without printing in correct order
• Not freeing nodes causing memory leaks
• Stopping after one pop