CS360 Midterm -- October 18, 1999. Question 1: Answer and Grading

10 points

The answer

Print_tree() should traverse the input tree. It should cast the val field of each node to a PTstruct, and then test the type field. If the type is zero, then it should print out the node's key, and fn.s. If the type is one, then it should cast fn.v to a JRB, traverse it and then print the original node's key with each sub-tree node's key:

void print_tree(JRB t)
{
  JRB tmp1, tmp2, subtree;
  PTstruct *p;

  jrb_traverse(tmp1, t) {
    p = (PTstruct *) tmp1->val.v;
    if (p->type == 0) {
      printf("%s %s\n", p->fn.s, tmp1->key.s);
    } else {
      subtree = (JRB) p->fn.v;
      jrb_traverse(tmp2, subtree) {
        printf("%s %s\n", tmp2->key.s, tmp1->key.s);
      }
    }
  }
}

Grading

Points were allocated as follows:

Overall structure: 1 point
Traversing the first tree: 1 point
Casting the val.v to a PTstruct *: 1 point
Printing the name when p->type = 0: 1 point (note, it did not matter if you printed the first name first, or the last name first, as long as you were consistent.
Printing the first name with fn.s: 1 point
Printing the last name with key.s: 1 point
Casting fn.v to a JRB if p->type = 1: 1 point
Traversing this second tree: 1 point
Printing the correct name for each subtree node: 1 point
Getting other details right: 1 point