Delete Node at End in DSA C - Execution Trace

Concept Flow - Delete Node at End

Start at head

↓

Check if list empty?

Yes→Nothing to delete, EXIT

No↓

Traverse to second last node

↓

Remove last node

↓

Update second last node's next to NULL

↓

Free last node memory

↓

End

Start from the head, check if list is empty, then move to the second last node, remove the last node by updating pointers and freeing memory.

Execution Sample

DSA C

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

void deleteAtEnd(Node** head) {
    if (*head == NULL) return;
    if ((*head)->next == NULL) {
        free(*head);
        *head = NULL;
        return;
    }
    Node* temp = *head;
    while (temp->next->next != NULL) temp = temp->next;
    free(temp->next);
    temp->next = NULL;
}

Deletes the last node of a singly linked list by traversing to the second last node and freeing the last node.

Execution Table

Step	Action	Pointer Positions	Condition	Result
1	Check if head is NULL	head -> Node1	head == NULL? No	Continue
2	Check if head->next is NULL	head -> Node1	head->next == NULL? No	Continue
3	Set temp = head	temp -> Node1	-	temp points to Node1
4	Check temp->next->next != NULL	temp -> Node1	Node2->next != NULL? Yes	Move temp to next
5	Move temp to next	temp -> Node2	-	temp points to Node2
6	Check temp->next->next != NULL	temp -> Node2	Node3->next != NULL? No	Stop loop
7	Free temp->next (Node3)	temp -> Node2	-	Node3 freed
8	Set temp->next = NULL	temp -> Node2	-	Node2->next = NULL
9	End	head -> Node1	-	Last node deleted, list ends at Node2

💡 Loop ends when temp->next->next is NULL, meaning temp is second last node.

Variable Tracker

Variable	Start	After Step 3	After Step 5	After Step 6	Final
head	Node1 -> Node2 -> Node3 -> NULL	Node1 -> Node2 -> Node3 -> NULL	Node1 -> Node2 -> Node3 -> NULL	Node1 -> Node2 -> Node3 -> NULL	Node1 -> Node2 -> NULL
temp	NULL	Node1	Node2	Node2	Node2

Key Moments - 3 Insights

Why do we check temp->next->next != NULL instead of temp->next != NULL in the loop?

Because we want temp to stop at the second last node, so that temp->next is the last node to delete. This is shown in execution_table rows 4-6.

What happens if the list has only one node?

The code frees the head node and sets head to NULL immediately (rows 1-2), avoiding traversal.

Why do we set temp->next = NULL after freeing the last node?

To update the second last node's next pointer to NULL, marking the new end of the list (row 8).

Visual Quiz - 3 Questions

Look at the execution_table, what node does temp point to after Step 5?

A Node1

B Node2

C Node3

D NULL

💡 Hint

Check the 'Pointer Positions' column at Step 5 in execution_table.

At which step does the loop stop traversing the list?

A Step 6

B Step 5

C Step 4

D Step 7

💡 Hint

Look for the step where condition temp->next->next != NULL becomes false in execution_table.

If the list had only one node, which step would free that node?

A Step 1

B Step 7

C Step 2

D Step 8

💡 Hint

Refer to the early check for single node list in execution_table rows 1-2.

Concept Snapshot

Delete Node at End in singly linked list:
- If list empty, do nothing.
- If only one node, free it and set head NULL.
- Else, traverse to second last node.
- Free last node and set second last's next to NULL.
- Updates list end safely.

Full Transcript

This visual execution traces deleting the last node in a singly linked list. It starts by checking if the list is empty or has only one node. If one node, it frees it and sets head to NULL. Otherwise, it traverses to the second last node by checking temp->next->next != NULL. Once at second last node, it frees the last node and sets second last's next pointer to NULL, marking the new end of the list. Variables head and temp change as shown. Key moments include why we check temp->next->next, handling single node list, and updating pointers after deletion. The quiz tests understanding of pointer positions and loop stopping conditions.

Practice

1. What is the main purpose of the Delete Node at End operation in a singly linked list?

easy

A. To reverse the linked list

B. To add a new node at the end of the linked list

C. To delete the first node of the linked list

D. To remove the last node from the linked list

Solution

1. Step 1: Understand the operation's goal

   The operation aims to remove the last node in the linked list, which is the node that points to null.

2. Step 2: Identify the correct description

   Removing the last node means deleting the node at the end, not adding or reversing nodes.

3. Final Answer:

   To remove the last node from the linked list -> Option D

4. Quick Check:

   Delete Node at End = Remove last node [OK]

Hint: Deleting at end means removing the last node [OK]

Common Mistakes:

• Confusing delete at end with delete at beginning
• Thinking it adds a node instead of deleting
• Mixing up with reversing the list

2. Which of the following C code snippets correctly deletes the last node of a singly linked list named head?

easy

A. if (head == NULL || head->next == NULL) return; Node* temp = head; while (temp->next->next != NULL) temp = temp->next; free(temp->next); temp->next = NULL;

B. Node* temp = head; while (temp->next != NULL) temp = temp->next; free(temp);

C. free(head); head = NULL;

D. Node* temp = head; while (temp->next != NULL) temp = temp->next; temp->next = NULL;

Solution

1. Step 1: Check for empty or single-node list and traverse to second last node

   if (head == NULL || head->next == NULL) return; Node* temp = head; while (temp->next->next != NULL) temp = temp->next; free(temp->next); temp->next = NULL; checks if list is empty or single node, then moves to the node before last using temp->next->next != NULL.

2. Step 2: Free last node and update pointer

   It frees the last node and sets temp->next to NULL, correctly deleting the last node.

3. Final Answer:

   if (head == NULL || head->next == NULL) return; Node* temp = head; while (temp->next->next != NULL) temp = temp->next; free(temp->next); temp->next = NULL; -> Option A

4. Quick Check:

   Traverse to second last, free last, set next NULL = if (head == NULL || head->next == NULL) return; Node* temp = head; while (temp->next->next != NULL) temp = temp->next; free(temp->next); temp->next = NULL; [OK]

Hint: Stop at second last node before deleting last [OK]

Common Mistakes:

• Freeing node without updating previous node's next pointer
• Not checking for empty list
• Freeing head without traversal

3. What will be the output linked list after deleting the last node from the list: 10 -> 20 -> 30 -> 40 -> NULL using the standard delete at end operation?

medium

A. 10 -> 20 -> 30 -> 40 -> NULL

B. 10 -> 20 -> 30 -> NULL

C. 20 -> 30 -> 40 -> NULL

D. 10 -> 20 -> NULL

Solution

1. Step 1: Identify the last node

   The last node is 40, which points to NULL.

2. Step 2: Remove the last node and update the second last node's next pointer

   After deletion, 30 becomes the last node and its next pointer is set to NULL.

3. Final Answer:

   10 -> 20 -> 30 -> NULL -> Option B

4. Quick Check:

   Last node 40 removed = 10 -> 20 -> 30 -> NULL [OK]

Hint: Last node removed, second last points to NULL [OK]

Common Mistakes:

• Not updating second last node's next pointer
• Deleting wrong node (like first node)
• Output list unchanged

4. Consider the following code snippet to delete the last node of a singly linked list. What is the error?

void deleteEnd(Node** head) {
  if (*head == NULL) return;
  Node* temp = *head;
  while (temp->next != NULL) {
    temp = temp->next;
  }
  free(temp);
  temp = NULL;
}

medium

A. It does not update the second last node's next pointer to NULL

B. It frees the head node instead of the last node

C. It causes a segmentation fault by accessing NULL pointer

D. It deletes the first node instead of the last

Solution

1. Step 1: Analyze traversal logic

   The loop moves temp to the last node (where temp->next is NULL).

2. Step 2: Check pointer update after freeing

   It frees the last node but does not update the second last node's next pointer to NULL, leaving a dangling pointer.

3. Final Answer:

   It does not update the second last node's next pointer to NULL -> Option A

4. Quick Check:

   Missing next pointer update causes dangling pointer [OK]

Hint: Always update second last node's next to NULL after free [OK]

Common Mistakes:

• Freeing node without updating previous node's next
• Not handling empty list
• Incorrect loop condition causing crash

5. You have a singly linked list with nodes: 5 -> 15 -> 25 -> 35 -> 45 -> NULL. Write the steps to delete the last node and then insert a new node with value 50 at the end. What will be the final list?

hard

A. 5 -> 15 -> 25 -> 35 -> NULL

B. 5 -> 15 -> 25 -> 35 -> 45 -> 50 -> NULL

C. 5 -> 15 -> 25 -> 35 -> 50 -> NULL

D. 50 -> 5 -> 15 -> 25 -> 35 -> NULL

Solution

1. Step 1: Delete the last node (45)

   Traverse to second last node (35), free last node (45), set 35's next to NULL.

2. Step 2: Insert new node with value 50 at end

   Traverse to last node (35), create new node with 50, set 35's next to new node, new node's next to NULL.

3. Final Answer:

   5 -> 15 -> 25 -> 35 -> 50 -> NULL -> Option C

4. Quick Check:

   Delete last, then add 50 at end = 5 -> 15 -> 25 -> 35 -> 50 -> NULL [OK]

Hint: Delete last node, then add new node at end [OK]

Common Mistakes:

• Not updating pointers correctly after deletion
• Appending new node without deleting last
• Inserting new node at wrong position