[Dev-C++] Copy Constructor for a BST Class 2

[Socha, D. F <dav...@us...>]

Would a copy constructor as such work:
=20

template <typename T> BST<T>::BST(const BST& origTree){     //copy =
constructor

	=09
		if(origTree.root){

				copyTree(origTree.root, root);

		}

		else{

				root =3D NULL;

		}

}


also, what do you mean by assigning to a non-empty tree.  I would think =
that the copy constructor would simply make a fresh (new) copy of the =
tree.  I dont understand why you would have to worry about releasing =
memory for a non empty tree

thanks,

David Socha

here is the copy tree code again


> > =
/************************************************************************=
****
> >
> > copyTree function
> >
> > Used by the copy constructor to copy the tree
> >
> > Sent: pointer to the original node, pointer to the node being copied =
to
> >
> > Returns: Nothing
> >
> > Works by: Creating a new node on the tree being copied, copy the =
data to the
> >
> > node, checks to see if a left node exists on the tree, if yes
> >
> > repeat the process using on the left node, if not, do nothing.  Then =
checks
> >
> > the right, if it exists repeat process on the right side, if not, do =
nothing
> >
> >  NOTE: This function is a recursive process
> >
> > =
*************************************************************************=
***/
> >
> >
> >
> > template<typename T> void BST<T>::copyTree(Node* origNode, Node* =
copyNode){
> >
> >      copyNode =3D new Node(origNode->data);
> >
> >
> >
> >      if(origNode->left){
> >
> >           copyTree(origNode->left, copyNode->left);
> >
> >      }
> >
> >
> >
> >      if(origNode->right){
> >
> >           copyTree(origNode->right, copyNode->right);
> >
> >      }
> >
> > }
> >
> > //End copyTree