Pop Using Linked List Node in DSA C - Execution Trace

Concept Flow - Pop Using Linked List Node

Start with head node

↓

Check if head is NULL?

Yes→List is empty, cannot pop

No↓

Store head node data

↓

Move head pointer to next node

↓

Free old head node

↓

Return stored data

↓

Done

This flow shows how popping removes the first node from the linked list by updating the head pointer and freeing the old node.

Execution Sample

DSA C

int pop(Node** head) {
  if (*head == NULL) return -1;
  Node* temp = *head;
  int val = temp->data;
  *head = temp->next;
  free(temp);
  return val;
}

This code removes the first node from the linked list, returns its data, and updates the head pointer.

Execution Table

Step	Operation	Nodes in List	Pointer Changes	Visual State
1	Start with list 10 -> 20 -> 30 -> NULL	10 -> 20 -> 30 -> NULL	head points to 10	head -> [10] -> [20] -> [30] -> NULL
2	Check if head is NULL	10 -> 20 -> 30 -> NULL	No change	head -> [10] -> [20] -> [30] -> NULL
3	Store data from head node	10 -> 20 -> 30 -> NULL	temp = head (10)	head -> [10] -> [20] -> [30] -> NULL
4	Move head to next node	10 -> 20 -> 30 -> NULL	head = head->next (20)	head -> [20] -> [30] -> NULL
5	Free old head node (10)	20 -> 30 -> NULL	temp freed	head -> [20] -> [30] -> NULL
6	Return stored data (10)	20 -> 30 -> NULL	No change	head -> [20] -> [30] -> NULL
7	Next pop call with list 20 -> 30 -> NULL	20 -> 30 -> NULL	head points to 20	head -> [20] -> [30] -> NULL
8	Pop again: store data 20	20 -> 30 -> NULL	temp = head (20)	head -> [20] -> [30] -> NULL
9	Move head to next node (30)	20 -> 30 -> NULL	head = head->next (30)	head -> [30] -> NULL
10	Free old head node (20)	30 -> NULL	temp freed	head -> [30] -> NULL
11	Return stored data (20)	30 -> NULL	No change	head -> [30] -> NULL
12	Pop again: store data 30	30 -> NULL	temp = head (30)	head -> [30] -> NULL
13	Move head to next node (NULL)	30 -> NULL	head = head->next (NULL)	head -> NULL
14	Free old head node (30)	NULL	temp freed	head -> NULL
15	Return stored data (30)	NULL	No change	head -> NULL
16	Pop on empty list	NULL	No change	head -> NULL
17	Return -1 (empty list)	NULL	No change	head -> NULL

💡 Execution stops when head becomes NULL and pop returns -1 indicating empty list.

Variable Tracker

Variable	Start	After Step 3	After Step 4	After Step 5	After Step 8	After Step 9	After Step 10	After Step 12	After Step 13	After Step 14	Final
head	points to [10]	points to [10]	points to [20]	points to [20]	points to [20]	points to [30]	points to [30]	points to [30]	NULL	NULL	NULL
temp	NULL	points to [10]	points to [10]	points to [10]	NULL	points to [20]	points to [20]	NULL	points to [30]	points to [30]	NULL
val	undefined	10	10	10	10	20	20	20	30	30	-1

Key Moments - 3 Insights

Why do we move the head pointer before freeing the old head node?

Because if we free the old head first, we lose access to the next node. Step 4 moves head to next before step 5 frees the old node, ensuring the list remains accessible.

What happens if we try to pop from an empty list?

As shown in steps 16 and 17, the function checks if head is NULL and returns -1 immediately, preventing errors.

Why do we store the data from the head node before moving the head pointer?

Because after moving head, the old node is freed and its data lost. Step 3 stores data first so it can be returned after the node is removed.

Visual Quiz - 3 Questions

Look at the execution table, what is the value of 'head' after step 4?

A Points to node with data 10

B Points to node with data 20

C NULL

D Points to node with data 30

💡 Hint

Check the 'Pointer Changes' and 'Visual State' columns at step 4.

At which step does the list become empty (head == NULL)?

A Step 13

B Step 14

C Step 15

D Step 16

💡 Hint

Look at the 'Pointer Changes' and 'Visual State' columns to see when head points to NULL.

If we skip freeing the old head node, what will happen in the execution table?

A Pointer changes will not update

B List will become empty immediately

C Visual state will still show nodes but memory leaks occur

D Stored data will be lost

💡 Hint

Refer to step 5 where freeing happens and the visual state remains the same but memory is freed.

Concept Snapshot

Pop Using Linked List Node:
- Remove first node (head) from list
- Store data from head
- Move head to next node
- Free old head node
- Return stored data
- If list empty, return -1

Full Transcript

This concept shows how to remove the first node from a linked list, called popping. We start by checking if the list is empty. If not, we save the data from the head node, move the head pointer to the next node, free the old head node to avoid memory leaks, and return the saved data. If the list is empty, we return -1. The execution table traces these steps with the list changing from 10->20->30 to empty after three pops.

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