Given the head of a doubly linked list and an integer key, delete every node whose value equals key — the key may appear any number of times, anywhere in the list — and return the head of the resulting list. This is a bypass problem, not a value-overwrite problem: a matching node is removed by redirecting the pointer in front of it to skip over it, not by copying values around. In a true doubly linked list, deleting a node also means pointing the node right after it back at the node right before it, so both directions of the chain stay consistent.