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206. Reverse Linked List

easyAsked at Atlassian

Reverse Linked List is Atlassian's pointer-manipulation warm-up. Given the head of a singly linked list, reverse the list and return the new head. The grading focus is correct three-pointer rewiring without dropping nodes or creating cycles.

By Alex Chen, Founder, InterviewChamp.AI · Last verified

Source citations

Public interview reports confirming this problem appears in Atlassian loops.

  • Glassdoor (2026-Q1)— Atlassian SWE-I / SWE-II phone-screen reports cite Reverse Linked List as a recurring pointer warm-up.
  • Reddit r/cscareerquestions (2025-10)— Atlassian onsite threads list Reverse Linked List as the 5-minute opener before harder list-manipulation problems.

Problem

Given the head of a singly linked list, reverse the list, and return the reversed list.

Constraints

  • The number of nodes in the list is the range [0, 5000].
  • -5000 <= Node.val <= 5000

Examples

Example 1

Input
head = [1,2,3,4,5]
Output
[5,4,3,2,1]

Example 2

Input
head = [1,2]
Output
[2,1]

Example 3

Input
head = []
Output
[]

Approaches

1. Iterative three-pointer (canonical)

Walk prev/curr/next. Save curr.next, flip curr.next to prev, advance prev and curr.

Time
O(n)
Space
O(1)
function reverseList(head) {
  let prev = null;
  let curr = head;
  while (curr) {
    const next = curr.next;
    curr.next = prev;
    prev = curr;
    curr = next;
  }
  return prev;
}

Tradeoff: The Atlassian-canonical answer. Six lines, no recursion overhead. Lead with this; mention recursion as a follow-up.

2. Recursive (returns new head from tail-up)

Recurse to the tail; on unwind set node.next.next = node and node.next = null.

Time
O(n)
Space
O(n) recursion
function reverseListRecursive(head) {
  if (!head || !head.next) return head;
  const newHead = reverseListRecursive(head.next);
  head.next.next = head;
  head.next = null;
  return newHead;
}

Tradeoff: Elegant but blows the JS stack at n=5000. Mention only after the iterative; explicitly call out the stack risk.

Atlassian-specific tips

Atlassian explicitly grades 'narrates pointer rewiring' on this problem. Talk through 'I save curr.next so I don't lose the tail, flip curr.next to prev, advance prev to curr, then advance curr to the saved next'. Many candidates write the loop silently and ship a subtle off-by-one — the narration is what catches it before the interviewer has to. After this they usually ask Reverse Nodes in k-Group (LeetCode 25) or Reverse Linked List II (range reverse).

Common mistakes

  • Forgetting to save curr.next before overwriting it — drops the rest of the list silently.
  • Returning curr instead of prev at the end — curr is null, prev is the new head.
  • Initializing prev to head instead of null — creates a cycle (last node points to head).

Follow-up questions

An interviewer at Atlassian may pivot to one of these next:

  • Reverse Linked List II (LeetCode 92) — reverse only positions m..n in the list.
  • Reverse Nodes in k-Group (LeetCode 25) — reverse the list k nodes at a time; final partial group stays unchanged.
  • Palindrome Linked List (LeetCode 234) — reverse the second half in place; compare halves.

Solve it now

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Output

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FAQ

Iterative or recursive at Atlassian?

Iterative. The recursive version is shorter on paper but adds O(n) stack space and risks overflow at n=5000 in JS. Atlassian's rubric scores the iterative version higher.

Why is it a 'foundational' problem?

Because the three-pointer pattern (prev / curr / next) is the load-bearing primitive behind half of the harder linked-list questions — copy-with-random-pointer, k-group reverse, list partition. Get it muscle-memory-clean.

Free learning resources

Curated free links for this problem.

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