Remove Element from the linked list - linked-list

Question

def removeKFromList(l, k):
pointer = l
while pointer:
if pointer.next and pointer.next.value == k:
pointer.next = pointer.next.next
else:
pointer = pointer.next
if l and l.value == k:
return l.next
else:
return l
In this code, why do I need to put
pointer = pointer.next
under else? Code does not work if I don't write this under else, but I don't see why.

Answer 1

First realise that pointer is intended to reference the node that precedes the one that might need to be deleted.
Now if we find that the successor of pointer must be deleted, then we get in the if block. In that case the removal of the next node will make the node that comes after that one, the new successor of pointer. As we want to make the check (for deletion) also for that new successor, we should not move pointer, but just leave it like that. In the next iteration we will then correctly determine whether that new successor node must be deleted or not.
If we would have moved pointer with pointer.next, then pointer would refer to the new successor, but it would not be checked for removal (in the next iteration). It would escape the removal check!
Here is a visualisation of what can go wrong when we do pointer = pointer.next in that case.
Input: l = [1,2,2,3], k = 2
l
pointer
↓
┌───────────┐ ┌───────────┐ ┌───────────┐ ┌───────────┐
│ value: 1 │ │ value: 2 │ │ value: 2 │ │ value: 3 │
│ next: ───────> │ next: ───────> │ next: ───────> │ next:None │
└───────────┘ └───────────┘ └───────────┘ └───────────┘
In the first iteration the condition pointer.next.value == k is true, and so the removal is performed with pointer.next = pointer.next.next, resulting in this situation:
l
pointer
↓ ┌────────────────┐
┌───────────┐ │ ┌───────────┐ │ ┌───────────┐ ┌───────────┐
│ value: 1 │ │ │ value: 2 │ └> │ value: 2 │ │ value: 3 │
│ next: ──────┘ │ next: ───────> │ next: ───────> │ next:None │
└───────────┘ └───────────┘ └───────────┘ └───────────┘
If now we would do pointer = pointer.next, then we get this:
l pointer
↓ ┌────────────────┐ ↓
┌───────────┐ │ ┌───────────┐ │ ┌───────────┐ ┌───────────┐
│ value: 1 │ │ │ value: 2 │ └> │ value: 2 │ │ value: 3 │
│ next: ──────┘ │ next: ───────> │ next: ───────> │ next:None │
└───────────┘ └───────────┘ └───────────┘ └───────────┘
...and in the next iteration the if condition will not be about that node, but about the next, the one with value 3, and so the second occurrence of 2 will not be deleted!