---
title: '206. Reverse Linked List'
description: Given the head of a singly linked list, reverse the list, and return the reversed list
icon: dot
topics:
  - { name: "Linked List", slug: "linked-list" }
  - { name: "Recursion", slug: "recursion" }
video: "G0_I-ZF0S38"
issue: "https://github.com/prdlk/leetcode/issues/77"
sidebar:
  label: 'Reverse Linked List'
  badge: 'Easy'
---

::::warning
A linked list can be reversed either iteratively or recursively. Could you implement both?
::::

### 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: `[]`

### Constraints:

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

## Approach

```mermaid
flowchart TD
  S(["reverseList(head)"]) --> I["prev = None, curr = head"]
  I --> W{"curr?"}
  W -- no --> E(["return prev — the old tail is the new head"])
  W -- yes --> A["next_ = curr.next — save it before overwriting"]
  A --> B["curr.next = prev — flip the link"]
  B --> C["prev = curr; curr = next_ — step forward"]
  C --> W
```

## Solution

```py
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
        prev = None
        curr = head
        while curr:
            next_ = curr.next
            curr.next = prev
            prev = curr
            curr = next_
        return prev
```