The add()method of the lock-free algorithm never finds a marked node with the same key. Can one modify the algorithm so that it will simply insert its new added object into the existing marked node...

The add()method of the lock-free algorithm never finds a marked node with the same key. Can one modify the algorithm so that it will simply insert its new added object into the existing marked node with the same key if such a node exists in the list, thus saving the need to insert a new node?