Exercise 3 (UPDATED): removing a node with two children
        There are two changes to this skeleton code, to handle the case that the
        right subtree of toRemove has no left subtree. The changes are:
            - added "if (leftMost->left != NULL)"
            - reordered the lines marked as 6, 7, and 8

        Do the following steps:
            - search for the leftmost node of the right subtree
            - remove the leftmost node (keep track of its children)
            - place the leftmost node in the same place as the node to remove
            - remove and delete the node to remove

        BTNode<Base> *parent, *toRemove;
        // ...
        // Assume that toRemove has two children.
        // Write the code that searches for the leftmost child of
        // the right subtree.


        BTNode<Base> *leftMost = _____________________________________________  // 1


                     *leftMostParent = ____________________________________     // 2

        if (leftMost->left != NULL) {


            while (___________________________________________) {               // 3


                ________________________________________________                // 4


                ________________________________________________                // 5
            }



            leftMostParent->left = ___________________________                  // 6


            leftMost->right = __________________________________                // 7
        }


        leftMost->left = ____________________________________                   // 8


        if (parent->left == toRemove) {


            ___________________________                                         // 9
        } else {


            ___________________________                                         // 10
        }


        toRemove->left = toRemove->right = _________________________________    // 11


        delete _______________________________                                  // 12