Leetcode - 206. Reverse Linked List

Easy

Dec 22, 2025

#linked-list

206. Reverse Linked List

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

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

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

Notes

Intuition

To reverse a linked list, we need to flip each node's pointer to point backward instead of forward. This can be done iteratively by maintaining references to the previous and current nodes.

Implementation

Use two pointers: curr starting at head and prev starting at None. For each node, save curr.next in a temp variable (we'll lose access to it otherwise), then reverse the link by setting curr.next = prev. Advance by setting prev = curr and curr = temp. When done, prev points to the new head.

Edge-cases

The order of operations is critical. Always save the next pointer before overwriting it. An empty list or single-node list works naturally—prev will be None or the single node respectively.

Time: O(n) — visit each node once
Space: O(1) — only using pointers

Complexity

Solution

from typing import Optional
from lcutils import ListNode

# 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]:
        curr = head
        prev = None
        while curr:
            temp = curr.next
            curr.next = prev
            prev = curr
            curr = temp
        return prev
        



if __name__ == "__main__":
    # Include one-off tests here or debugging logic that can be run by running this file
    # e.g. print(solution.two_sum([1, 2, 3, 4], 3))
    solution = Solution()