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

Concept Flow - Reverse a Singly Linked List Iterative

Start with head node

↓

Initialize prev = None, current = head

↓

While current is not None

↓

Save next node: next_node = current.next

↓

Reverse pointer: current.next = prev

↓

Move prev to current

↓

Move current to next_node

↓

Repeat loop

↓

When current is None, prev is new head

↓

Return prev as reversed list head

We start from the head and move through the list, reversing each pointer one by one until we reach the end. The prev pointer tracks the reversed part.

Execution Sample

DSA Python

def reverse_linked_list(head):
    prev = None
    current = head
    while current:
        next_node = current.next
        current.next = prev
        prev = current
        current = next_node
    return prev

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

Execution Table

Step	Operation	Nodes in List	Pointer Changes	Visual State
1	Initialize prev=None, current=head (1)	1 -> 2 -> 3 -> 4 -> None	prev=None, current=1	1 -> 2 -> 3 -> 4 -> None
2	Save next_node = current.next (2)	1 -> 2 -> 3 -> 4 -> None	next_node=2	1 -> 2 -> 3 -> 4 -> None
3	Reverse current.next to prev (3)	1 -> 2 -> 3 -> 4 -> None	1.next=None	1 -> None, 2 -> 3 -> 4 -> None
4	Move prev to current (4)	1 -> None, 2 -> 3 -> 4 -> None	prev=1	1 -> None, 2 -> 3 -> 4 -> None
5	Move current to next_node (5)	1 -> None, 2 -> 3 -> 4 -> None	current=2	1 -> None, 2 -> 3 -> 4 -> None
6	Save next_node = current.next (6)	1 -> None, 2 -> 3 -> 4 -> None	next_node=3	1 -> None, 2 -> 3 -> 4 -> None
7	Reverse current.next to prev (7)	1 -> None, 2 -> 3 -> 4 -> None	2.next=1	2 -> 1 -> None, 3 -> 4 -> None
8	Move prev to current (8)	2 -> 1 -> None, 3 -> 4 -> None	prev=2	2 -> 1 -> None, 3 -> 4 -> None
9	Move current to next_node (9)	2 -> 1 -> None, 3 -> 4 -> None	current=3	2 -> 1 -> None, 3 -> 4 -> None
10	Save next_node = current.next (10)	2 -> 1 -> None, 3 -> 4 -> None	next_node=4	2 -> 1 -> None, 3 -> 4 -> None
11	Reverse current.next to prev (11)	2 -> 1 -> None, 3 -> 4 -> None	3.next=2	3 -> 2 -> 1 -> None, 4 -> None
12	Move prev to current (12)	3 -> 2 -> 1 -> None, 4 -> None	prev=3	3 -> 2 -> 1 -> None, 4 -> None
13	Move current to next_node (13)	3 -> 2 -> 1 -> None, 4 -> None	current=4	3 -> 2 -> 1 -> None, 4 -> None
14	Save next_node = current.next (14)	3 -> 2 -> 1 -> None, 4 -> None	next_node=None	3 -> 2 -> 1 -> None, 4 -> None
15	Reverse current.next to prev (15)	3 -> 2 -> 1 -> None, 4 -> None	4.next=3	4 -> 3 -> 2 -> 1 -> None
16	Move prev to current (16)	4 -> 3 -> 2 -> 1 -> None	prev=4	4 -> 3 -> 2 -> 1 -> None
17	Move current to next_node (17)	4 -> 3 -> 2 -> 1 -> None	current=None	4 -> 3 -> 2 -> 1 -> None
18	Loop ends, return prev as new head (18)	4 -> 3 -> 2 -> 1 -> None	return prev=4	4 -> 3 -> 2 -> 1 -> None

💡 current becomes None at step 17, loop ends, prev points to new head

Variable Tracker

Variable	Start	After 1	After 2	After 3	After 4	Final
prev	None	1	2	3	4	4
current	1	2	3	4	None	None
next_node	None	2	3	4	None	None

Key Moments - 3 Insights

Why do we set current.next = prev instead of prev.next = current?

Because we reverse the link from current node to previous node. Setting current.next = prev changes the direction of the pointer. prev.next = current would create a cycle. See execution_table rows 3, 7, 11, 15.

Why do we need to save next_node before reversing the pointer?

Because once we do current.next = prev, we lose the original next node. Saving next_node before reversing lets us continue traversal. See execution_table rows 2, 6, 10, 14.

What happens when current becomes None?

It means we reached the end of the original list. The prev pointer now points to the new head of the reversed list. The loop ends and we return prev. See execution_table rows 17 and 18.

Visual Quiz - 3 Questions

Look at the execution_table at step 7, what is the visual state of the list?

A 2 -> 1 -> None, 3 -> 4 -> None

B 3 -> 2 -> 1 -> None, 4 -> None

C 1 -> None, 2 -> 3 -> 4 -> None

D 4 -> 3 -> 2 -> 1 -> None

💡 Hint

Check the Visual State column at step 7 in execution_table

At which step does current become None, ending the loop?

A Step 18

B Step 16

C Step 17

D Step 15

💡 Hint

Look at Pointer Changes column for current value in execution_table

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

A The list remains unchanged

B The reversed list will only contain the first node

C The list is reversed correctly

D The reversed list will be empty

💡 Hint

Check variable_tracker for prev changes and how it affects the Visual State

Concept Snapshot

Reverse a singly linked list iteratively:
- Use three pointers: prev=None, current=head, next_node
- Loop while current is not None:
  * Save next_node = current.next
  * Reverse current.next to prev
  * Move prev to current
  * Move current to next_node
- Return prev as new head
This reverses links one by one until list end.

Full Transcript

This concept shows how to reverse a singly linked list using an iterative method. We start with two pointers: prev set to None and current set to the head of the list. In each step, we save the next node to continue traversal after reversing the pointer. Then we reverse the current node's next pointer to point to prev, effectively flipping the link direction. We move prev to current and current to the saved next node. This repeats until current becomes None, meaning we reached the end. At that point, prev points to the new head of the reversed list, which we return. The execution table tracks each step, showing how nodes and pointers change, and the variable tracker shows how prev, current, and next_node update. Key moments clarify why we save next_node before reversing and why we set current.next to prev. The visual quiz tests understanding of list states at different steps and the importance of updating pointers correctly.