Reverse a Singly Linked List Iterative in DSA Python - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main idea behind reversing a singly linked list iteratively?

We change the direction of the links one by one, so each node points to its previous node instead of the next. We use three pointers: previous, current, and next to keep track while moving through the list.

beginner

In the iterative reversal of a singly linked list, what role does the 'previous' pointer play?

The 'previous' pointer keeps track of the node that should come after the current node once the link is reversed. It starts as None and moves forward as we reverse links.

intermediate

Why do we need to save the 'next' node before changing the current node's next pointer during reversal?

Because once we change the current node's next pointer to point backward, we lose the original forward link. Saving 'next' lets us continue traversing the list without losing nodes.

beginner

What is the time complexity of reversing a singly linked list iteratively?

The time complexity is O(n), where n is the number of nodes, because we visit each node exactly once to reverse the links.

beginner

Show the final state of the linked list after reversing 1 -> 2 -> 3 -> null iteratively.

After reversal, the linked list becomes 3 -> 2 -> 1 -> null.

Which pointer should initially be set to None when reversing a singly linked list iteratively?

A previous

B current

C next

D head

The 'previous' pointer starts as None because the new tail of the reversed list will point to None.

What must you do before changing the current node's next pointer during reversal?

A Delete the current node

B Set previous to None

C Move current to previous

D Save the next node

Saving the next node is necessary to continue traversal after changing the current node's next pointer.

What is the new head of the list after iterative reversal?

A The original tail node

B The original head node

C The middle node

D None

After reversal, the original tail becomes the new head.

How many times do we visit each node during iterative reversal?

A Twice

B Once

C Three times

D Zero times

Each node is visited exactly once to reverse the links.

What happens if you forget to update the 'previous' pointer during iteration?

A The list will become empty

B The list will reverse correctly

C The list will not reverse correctly

D The list will double in size

Not updating 'previous' breaks the chain of reversed links, so reversal fails.

Explain step-by-step how to reverse a singly linked list iteratively.

Think about how pointers move and change direction.

Describe why saving the next node before reversing the link is crucial.

Imagine losing your map while walking backwards.