Task $3\mathrm{.}4$:

void BST::remove$($const Animal$*$ a$)$

Remove an animal from the BST$.$ The steps are similar to the lecture notes with a few differences:

If the node with the "same" value is found, remove it from the linked list.

If the linked list becomes empty, the BSTnode needs to be removed from the BST$.$ This means:

If this node has $2$ children, "move" the linked list from the minimum$-$value node of the right BST to the current node, and remove the old minimum$-$value node from the right BST$.$

Else, remove and deallocate this node similar to the lecture example.