9.3.3 Symmetric Case |
345. <Right-side rebalancing after TRB insertion 345> = struct trb_node *y = pa[k - 2]->trb_link[0]; if (pa[k - 2]->trb_tag[0] == TRB_CHILD && y->trb_color == TRB_RED) {
<Case 1 in right-side TRB insertion rebalancing 346>
} else
{ struct trb_node *x; if (da[k - 1] == 1) y = pa[k - 1]; else
{
<Case 3 in right-side TRB insertion rebalancing 348>
} <Case 2 in right-side TRB insertion rebalancing 347> break; }
346. <Case 1 in right-side TRB insertion rebalancing 346> = <Case 1 in right-side RB insertion rebalancing; rb => trb 207>
347. <Case 2 in right-side TRB insertion rebalancing 347> = <Case 2 in right-side RB insertion rebalancing; rb => trb 208> if (y->trb_tag[0] == TRB_THREAD)
{ y->trb_tag[0] = TRB_CHILD; x->trb_tag[1] = TRB_THREAD; x->trb_link[1] = y; }
348. <Case 3 in right-side TRB insertion rebalancing 348> = <Case 3 in right-side RB insertion rebalancing; rb => trb 209> if (y->trb_tag[1] == TRB_THREAD)
{ y->trb_tag[1] = TRB_CHILD; x->trb_tag[0] = TRB_THREAD; x->trb_link[0] = y; }
Exercises:
1. It could be argued that the algorithm here is "impure" because it uses a stack, when elimination of the need for a stack is one of the reasons originally given for using threaded trees. Write a version of trb_probe() that avoids the use of a stack. You can use find_parent() from <Find parent of a TBST node 327> as a substitute. [answer]
9.3.2 Step 3: Rebalance | 9.3 Insertion | 9.4 Deletion |