206. Reverse Linked List §

Date: 2026-03-16 Difficulty: Easy Topics: Linked List, Recursion Link: https://leetcode.com/problems/reverse-linked-list/

Final Solution §

class Solution:
    def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
        if not head or not head.next:
            return head
        newHead = self.reverseList(head.next)
        head.next.next = head
        head.next = None
        return newHead

Initial Issues §

• Struggled significantly with the recursive approach
• Specifically confused by the pointer manipulation: head.next.next = head and head.next = None
• Hard to visualize what’s happening as recursion unwinds

Key Learnings §

• The recursive solution works by recursing to the end of the list first, then reversing pointers on the way back up the call stack
• newHead is always the last node (found at base case) and gets passed all the way back unchanged
• The key insight: when recursion unwinds at node head, head.next still points to the next node. So head.next.next = head makes that next node point back to head (reversing the link)
• head.next = None breaks the old forward link to avoid cycles

Nuances to Remember §

• Walk through example: 1->2->3->None
  • Base case returns 3 as newHead
  • At head=2: head.next is 3, so 3.next = 2 (now 3->2), then 2.next = None
  • At head=1: head.next is 2, so 2.next = 1 (now 2->1), then 1.next = None
  • Result: 3->2->1->None, newHead is still 3
• The iterative approach (prev/curr/next pointers) is more intuitive but recursive is important to understand for interview follow-ups
• head.next.next is NOT double-jumping — it’s accessing the next pointer OF the node that head.next points to