Right-Looking Deletion in a RTBST

Our usual algorithm for deletion looks at the right subtree of the node to be deleted, so we call it “right-looking.” The outline for this kind of deletion is the same as in TBST deletion (see Deleting from a TBST):

383. <Step 2: Delete RTBST node, right-looking 383> =
if (p->rtbst_rtag == RTBST_THREAD) 
  {
    if (p->rtbst_link[0] != NULL)
      { 
        <Case 1 in right-looking RTBST deletion 384> 
      }
    else 
      { 
        <Case 2 in right-looking RTBST deletion 385> 
      }
  } 
else 
  {
    struct rtbst_node *r = p->rtbst_link[1];
    if (r->rtbst_link[0] == NULL)
      { 
        <Case 3 in right-looking RTBST deletion 386> 
      }
    else 
      { 
        <Case 4 in right-looking RTBST deletion 387> 
      }
  }

Each of the four cases, presented below, is closely analogous to the same case in TBST deletion.

Case 1: p has a right thread and a left child

In this case, node p has a right thread and a left child. As in a TBST, this means that after deleting p we must update the right thread in p's former left subtree to point to p's replacement. The only difference from <Case 1 in TBST deletion 260> is in structure members:

384. <Case 1 in right-looking RTBST deletion 384> =
struct rtbst_node *t = p->rtbst_link[0];
while (t->rtbst_rtag == RTBST_CHILD)
  t = t->rtbst_link[1];
t->rtbst_link[1] = p->rtbst_link[1];
q->rtbst_link[dir] = p->rtbst_link[0];

This code is included in 383.

Case 2: p has a right thread and no left child

If node p is a leaf, then there are two subcases, according to whether p is a left child or a right child of its parent q. If dir is 0, then p is a left child and the pointer from its parent must be set to NULL. If dir is 1, then p is a right child and the link from its parent must be changed to a thread to its successor.

In either of these cases we must set q->rtbst_link[dir]: if dir is 0, we set it to NULL, otherwise dir is 1 and we set it to p->rtbst_link[1]. However, we know that p->rtbst_link[0] is NULL, because p is a leaf, so we can instead unconditionally assign p->rtbst_link[dir]. In addition, if dir is 1, then we must tag q's right link as a thread.

If q is the pseudo-root, then dir is 0 and everything works out fine with no need for a special case.

385. <Case 2 in right-looking RTBST deletion 385> =
q->rtbst_link[dir] = p->rtbst_link[dir];
if (dir == 1)
  q->rtbst_rtag = RTBST_THREAD;

This code is included in 383.

Case 3: p's right child has no left child

Code for this case, where p has a right child r that itself has no left child, is almost identical to <Case 3 in TBST deletion 262>. There is no left tag to copy, but it is still necessary to chase down the right thread in r's new left subtree (the same as p's former left subtree):

386. <Case 3 in right-looking RTBST deletion 386> =
r->rtbst_link[0] = p->rtbst_link[0];
if (r->rtbst_link[0] != NULL) 
  {
    struct rtbst_node *t = r->rtbst_link[0];
    while (t->rtbst_rtag == RTBST_CHILD)
      t = t->rtbst_link[1];
    t->rtbst_link[1] = r;
  }
q->rtbst_link[dir] = r;

This code is included in 383.

Case 4: p's right child has a left child

Code for case 4, the most general case, is very similar to <Case 4 in TBST deletion 263>. The only notable difference is in the subcase where s has a right thread: in that case we just set r's left link to NULL instead of having to set it up as a thread.

387. <Case 4 in right-looking RTBST deletion 387> =
struct rtbst_node *s;

for (;;) 
  {
    s = r->rtbst_link[0];
    if (s->rtbst_link[0] == NULL)
      break;

    r = s;
  }

if (s->rtbst_rtag == RTBST_CHILD)
  r->rtbst_link[0] = s->rtbst_link[1];
else 
  r->rtbst_link[0] = NULL;

s->rtbst_link[0] = p->rtbst_link[0];
if (p->rtbst_link[0] != NULL) 
  {
    struct rtbst_node *t = p->rtbst_link[0];
    while (t->rtbst_rtag == RTBST_CHILD)
      t = t->rtbst_link[1];
    t->rtbst_link[1] = s;
  }

s->rtbst_link[1] = p->rtbst_link[1];
s->rtbst_rtag = RTBST_CHILD;

q->rtbst_link[dir] = s;

This code is included in 383.

Exercises:

1. Rewrite <Case 4 in right-looking RTBST deletion 387> to replace the deleted node's rtavl_data by its successor, then delete the successor, instead of shuffling pointers. (Refer back to Exercise 4.8-3 for an explanation of why this approach cannot be used in libavl.) [answer]