Reverse a Singly Linked List Iterative in DSA Python

Mental Model

We change the direction of each link one by one so the list points backward instead of forward.

Analogy: Imagine a line of people holding hands forward. We ask each person to turn around and hold the hand of the person behind them instead.

head -> 1 -> 2 -> 3 -> null

Dry Run Walkthrough

Input: list: 1 -> 2 -> 3 -> null

Goal: Reverse the list so it becomes 3 -> 2 -> 1 -> null

Step 1: Initialize prev to null and curr to head (node 1)

prev = null, curr = 1 -> 2 -> 3 -> null

Why: We start with prev as null because the new tail will point to null

Step 2: Save next node (2), reverse curr's pointer to prev (null), move prev to curr (1), move curr to next (2)

1 -> null [prev]
2 -> 3 -> null [curr]

Why: We reverse the first link and move forward to continue reversing

Step 3: Save next node (3), reverse curr's pointer to prev (1), move prev to curr (2), move curr to next (3)

2 -> 1 -> null [prev]
3 -> null [curr]

Why: We reverse the second link and move forward

Step 4: Save next node (null), reverse curr's pointer to prev (2), move prev to curr (3), move curr to next (null)

3 -> 2 -> 1 -> null [prev]
curr = null

Why: We reverse the last link; curr is now null so we stop

Result:

3 -> 2 -> 1 -> null

Annotated Code

DSA Python

class Node:
    def __init__(self, val):
        self.val = val
        self.next = None

def reverse_linked_list(head):
    prev = None
    curr = head
    while curr is not None:
        next_node = curr.next  # save next node
        curr.next = prev       # reverse pointer
        prev = curr            # move prev forward
        curr = next_node       # move curr forward
    return prev

def print_list(head):
    curr = head
    result = []
    while curr:
        result.append(str(curr.val))
        curr = curr.next
    print(' -> '.join(result) + ' -> null')

# Driver code
head = Node(1)
head.next = Node(2)
head.next.next = Node(3)

print_list(head)
reversed_head = reverse_linked_list(head)
print_list(reversed_head)

while curr is not None:

loop until we reach the end of the list

next_node = curr.next # save next node

keep track of next node before changing pointer

curr.next = prev # reverse pointer

reverse current node's pointer to previous node

prev = curr # move prev forward

advance prev to current node

curr = next_node # move curr forward

advance curr to next node to continue

return prev

prev is new head after full reversal

Output

1 -> 2 -> 3 -> null

3 -> 2 -> 1 -> null

Complexity Analysis

Time: O(n) because we visit each node once to reverse its pointer

Space: O(1) because we only use a few extra variables, no extra list

vs Alternative: Compared to recursive reversal which uses O(n) space on call stack, iterative uses constant space

Edge Cases

empty list (head is null)

function returns null immediately, no reversal needed

DSA Python

while curr is not None:

single node list

pointer reversal does nothing visible, returns same single node

DSA Python

while curr is not None:

When to Use This Pattern

When you need to reverse the order of a singly linked list, use the iterative reversal pattern by changing pointers one by one.

Common Mistakes

Mistake: Not saving the next node before reversing pointer causes loss of access to rest of list
Fix: Always save curr.next in a variable before changing curr.next

Mistake: Returning head instead of prev after reversal
Fix: Return prev because it points to new head after reversal

Summary

Reverses a singly linked list by changing each node's pointer to the previous node.

Use when you want to reverse the order of nodes in a singly linked list efficiently.

The key is to save the next node before reversing the pointer to avoid losing the rest of the list.