# Reverse Linked List II LeetCode Solution | Easy Approach

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Given the `head` of a singly linked list and two integers `left` and `right` where `left <= right`, reverse the nodes of the list from position `left` to position `right`, and return the reversed list.

Example 1:

```Input: head = [1,2,3,4,5], left = 2, right = 4
Output: [1,4,3,2,5]
```

Example 2:

```Input: head = , left = 1, right = 1
Output: 
```

Constraints:

• The number of nodes in the list is `n`.
• `1 <= n <= 500`
• `-500 <= Node.val <= 500`
• `1 <= left <= right <= n`

Time: O(n)
Space: O(n)

### C++

``````class Solution {
public:
ListNode* reverseBetween(ListNode* head, int left, int right) {
if (left == 1)

}

private:
ListNode* reverseN(ListNode* head, int n) {
if (n == 1)

}
};
``````

### Java

``````

class Solution {
public ListNode reverseBetween(ListNode head, int left, int right) {
if (left == 1)

}

private ListNode reverseN(ListNode head, int n) {
if (n == 1)

}
}
``````

### Python

`````` class Solution:
def reverseBetween(self, head: Optional[ListNode], left: int, right: int) -> Optional[ListNode]:
if left == 1:

def reverseN(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:
if n == 1: