The iterative version came out fine, three pointers, prev starts null, walk through the list swapping next pointers.
Start by clarifying the problem and edge cases, then explain the iterative approach using three pointers (prev, curr, next) to reverse the links in one pass. Follow with the recursive approach, highlighting the base case and the recursive step that reverses the rest of the list and adjusts pointers. Finally, compare time and space complexities and discuss trade-offs.
Pro tip: Mention that while recursion is elegant, it uses O(n) stack space, so iterative is preferred for large lists to avoid stack overflow. Also, write clean code with meaningful variable names and handle edge cases like empty list or single node.
Confirm the function signature, input/output, and constraints. Discuss edge cases: empty list, single node, and list with multiple nodes.
Describe using three pointers: prev (initially null), curr (head), and next. Iterate through the list, reversing the link at each step, and finally return prev as the new head.
Define base case: if head is null or head.next is null, return head. Recursively reverse the rest, then set head.next.next = head and head.next = null, returning the new head from recursion.
State that both approaches run in O(n) time. Iterative uses O(1) space, while recursive uses O(n) space due to call stack. Discuss when to prefer each.
Walk through a small example (e.g., 1->2->3->null) for both approaches to verify correctness and demonstrate understanding.
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