Swap Nodes in Pairs

Given a linked list, swap every two adjacent nodes and return its head. You must solve the problem without modifying the values in the list's nodes (i.e., only nodes themselves may be changed).

Example: 1->2->3->4 becomes 2->1->4->3.

We could just iterate through the list and swap curr.val with curr.next.val. This is easy and perfectly fine in many real-world scenarios, but interviewers explicitly ban it to test your pointer manipulation skills.

To swap nodes A and B (prev -> A -> B -> next), we need to change 3 pointers:

1. prev.next should point to B.
2. A.next should point to next.
3. B.next should point to A.

Recursive: Swap the first two nodes, then recursively call the function for the rest of the list.
Iterative: Use a dummy node and a prev pointer to handle swaps in a loop.

Watch how the pointers are re-wired to swap nodes.