8.5.4 Step 4: Rebalance

Rebalancing after deletion in a TAVL tree divides into three cases. The first of these is analogous to case 1 in unthreaded AVL deletion, the other two to case 2 (see Inserting into a TBST). The cases are distinguished, as usual, based on the balance factor of right child x of the node y at which rebalancing occurs:

321. <Step 4: Rebalance after TAVL deletion 321> =
struct tavl_node *x = y->tavl_link[1];

assert (x != NULL);
if (x->tavl_balance == -1) 
  {
    <Rebalance for - balance factor after TAVL deletion in left subtree 322>
  } 
else 
  {
    q->tavl_link[dir] = x;

    if (x->tavl_balance == 0) 
      {
        <Rebalance for 0 balance factor after TAVL deletion in left subtree 323>
        break;
      } 
    else /* x->tavl_balance == +1 */ 
      {
        <Rebalance for + balance factor after TAVL deletion in left subtree 324>
      }
  }

This code is included in 320.

Case 1: x has - balance factor

This case is just like case 2 in TAVL insertion. In fact, we can even reuse the code:

322. <Rebalance for - balance factor after TAVL deletion in left subtree 322> =
struct tavl_node *w;

<Rebalance for - balance factor in TAVL insertion in right subtree 312>
q->tavl_link[dir] = w;

This code is included in 321.

Case 2: x has 0 balance factor

If x has a 0 balance factor, then we perform a left rotation at y. The transformation looks like this, with subtree heights listed under their labels:

Subtree b is taller than subtree a, so even if h takes its minimum value of 1, then subtree b has height h == 1 and, therefore, it must contain at least one node and there is no need to do any checking for threads. The code is simple:

323. <Rebalance for 0 balance factor after TAVL deletion in left subtree 323> =
y->tavl_link[1] = x->tavl_link[0];
x->tavl_link[0] = y;
x->tavl_balance = -1;
y->tavl_balance = +1;

This code is included in 321 and 445.

Case 3: x has + balance factor

If x has a + balance factor, we perform a left rotation at y, same as for case 2, and the transformation looks like this:

One difference from case 2 is in the resulting balance factors. The other is that if h == 1, then subtrees a and b have height h - 1 == 0, so a and b may actually be threads. In that case, the transformation must be done this way:

This code handles both possibilities:

324. <Rebalance for + balance factor after TAVL deletion in left subtree 324> =
if (x->tavl_tag[0] == TAVL_CHILD)
  y->tavl_link[1] = x->tavl_link[0];
else 
  {
    y->tavl_tag[1] = TAVL_THREAD;
    x->tavl_tag[0] = TAVL_CHILD;
  }
x->tavl_link[0] = y;  
y->tavl_balance = x->tavl_balance = 0;

This code is included in 321.