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apps/docs/content/linked-list/206-reverse-linked-list.mdx

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+---
+title: '206. Reverse Linked List'
+description: Given the head of a singly linked list, reverse the list, and return the reversed list
+sidebar:
+  label: 'Reverse Linked List'
+  badge: 'Easy'
+---
+
+<Badge variant="accent">Linked List</Badge>
+
+::::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
+```
+
+## Explanation
+
+<YouTube id="G0_I-ZF0S38" title="Reverse Linked List - Iterative AND Recursive - Leetcode 206 - Python" />