Reverse a Singly Linked List Iterative in DSA C - Execution Trace

Concept Flow - Reverse a Singly Linked List Iterative

Start with head node

↓

Initialize prev = NULL, current = head

↓

While current != NULL

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Save next = current->next

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Reverse pointer: current->next = prev

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Move prev = current

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Move current = next

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Repeat loop

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Loop ends when current == NULL

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Return prev as new head

We start from the head and move through the list, reversing each node's pointer to point backward, until we reach the end. The prev pointer becomes the new head.

Execution Sample

DSA C

Node* reverseList(Node* head) {
  Node* prev = NULL;
  Node* current = head;
  while (current != NULL) {
    Node* next = current->next;
    current->next = prev;
    prev = current;
    current = next;
  }
  return prev;
}

This code reverses a singly linked list by changing the next pointers of each node iteratively.

Execution Table

Step	Operation	Nodes in List	Pointer Changes	Visual State
0	Initialize prev=NULL, current=head (1)	1 -> 2 -> 3 -> NULL	prev=NULL, current=1	1 -> 2 -> 3 -> NULL
1	Save next = current->next (2)	1 -> 2 -> 3 -> NULL	next=2	1 -> 2 -> 3 -> NULL
2	Reverse current->next to prev (NULL)	1 -> 2 -> 3 -> NULL	1->next=NULL	1 -> NULL 2 -> 3 -> NULL
3	Move prev = current (1)	1 -> 2 -> 3 -> NULL	prev=1	1 -> NULL 2 -> 3 -> NULL
4	Move current = next (2)	1 -> 2 -> 3 -> NULL	current=2	1 -> NULL 2 -> 3 -> NULL
5	Save next = current->next (3)	1 -> NULL 2 -> 3 -> NULL	next=3	1 -> NULL 2 -> 3 -> NULL
6	Reverse current->next to prev (1)	1 -> NULL 2 -> 3 -> NULL	2->next=1	2 -> 1 -> NULL 3 -> NULL
7	Move prev = current (2)	2 -> 1 -> NULL 3 -> NULL	prev=2	2 -> 1 -> NULL 3 -> NULL
8	Move current = next (3)	2 -> 1 -> NULL 3 -> NULL	current=3	2 -> 1 -> NULL 3 -> NULL
9	Save next = current->next (NULL)	2 -> 1 -> NULL 3 -> NULL	next=NULL	2 -> 1 -> NULL 3 -> NULL
10	Reverse current->next to prev (2)	2 -> 1 -> NULL 3 -> NULL	3->next=2	3 -> 2 -> 1 -> NULL
11	Move prev = current (3)	3 -> 2 -> 1 -> NULL	prev=3	3 -> 2 -> 1 -> NULL
12	Move current = next (NULL)	3 -> 2 -> 1 -> NULL	current=NULL	3 -> 2 -> 1 -> NULL
13	Loop ends, return prev as new head (3)	3 -> 2 -> 1 -> NULL	return prev	3 -> 2 -> 1 -> NULL

💡 current becomes NULL at step 12, loop ends, prev points to new head

Variable Tracker

Variable	Start	After 1	After 2	After 3	After 4	After 5	Final
prev	NULL	1	2	3	3	3	3
current	1	2	3	NULL	NULL	NULL	NULL
next	undefined	2	3	NULL	NULL	NULL	NULL

Key Moments - 3 Insights

Why do we need to save 'next' before reversing the pointer?

Because once we reverse current->next to prev, we lose the original next node. Saving 'next' first (see step 1 and 5) lets us continue traversing the list.

Why does the loop stop when current becomes NULL?

The loop condition is while(current != NULL). When current is NULL (step 12), it means we reached past the last node, so reversal is complete.

What does 'prev' point to at the end?

At the end (step 13), 'prev' points to the new head of the reversed list, which was the last node before reversal.

Visual Quiz - 3 Questions

Look at the execution table at step 6, what does node 2's next pointer point to?

A NULL

B Node 3

C Node 1

D Node 2 itself

💡 Hint

Check the 'Pointer Changes' column at step 6 where 2->next is set to 1.

At which step does the 'current' pointer become NULL?

A Step 12

B Step 10

C Step 13

D Step 9

💡 Hint

Look at the 'current' variable in variable_tracker and execution_table rows.

If we forget to update 'prev = current' inside the loop, what happens to the final list?

A List remains unchanged

B List is partially reversed or broken

C List becomes empty

D List is reversed correctly

💡 Hint

Refer to the importance of 'prev' updates in the execution_table steps 3, 7, and 11.

Concept Snapshot

Reverse a singly linked list iteratively:
- Use three pointers: prev (NULL), current (head), next (temp)
- Loop while current != NULL:
  * Save next = current->next
  * Reverse current->next = prev
  * Move prev = current
  * Move current = next
- Return prev as new head
This reverses links one by one, changing direction.

Full Transcript

This visualization shows how to reverse a singly linked list using an iterative method. We start with three pointers: prev set to NULL, current set to the head of the list, and next used to temporarily store the next node. In each step, we save the next node, reverse the current node's pointer to point to prev, then move prev and current forward. This continues until current becomes NULL, meaning we have processed all nodes. The prev pointer then points to the new head of the reversed list. The execution table tracks each step, showing how pointers change and how the list nodes rearrange visually. Key moments clarify why saving next is necessary before reversing, why the loop stops when current is NULL, and what prev points to at the end. The quiz tests understanding of pointer changes and loop termination. This method efficiently reverses the list in-place without extra memory.