LeetCode 0206 - Reverse Linked List

• Difficulty: Easy
• Topics: Linked List, Recursion

---

Optimal Approach: Iterative Reversal

Intuition

Maintain three pointers: prev, curr, and next_node. At each step, flip curr.next to point to prev.

Code Implementation

# 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: ListNode) -> ListNode:
        prev = None
        curr = head
        while curr:
            next_node = curr.next
            curr.next = prev
            prev = curr
            curr = next_node
        return prev

Complexity Analysis

• Time Complexity: $\mathcal{O}(N)$
• Space Complexity: $\mathcal{O}(1)$