P4 #6
Description
In this deletion case, the node to be deleted is just mimicking the next node and is not actually being dereferenced, right? So instead of deleting node2, we're really just skipping over node3 and using its value and next reference. Would this mean that node2 will remain in memory until the program finishes? You'd need to change the next reference of node1 to node3 in order for node2 to be cleaned up sooner. Please let me know if I'm mistaken here.