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;
}
```

This code is included in 340.

```346. <Case 1 in right-side TRB insertion rebalancing 346> =
<Case 1 in right-side RB insertion rebalancing; rb => trb 207>
```

This code is included in 345.

```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;
}
```

This code is included in 345.

```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;
}
```

This code is included in 345.

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]