Initial Black Insertion in an RB Tree - GNU libavl 2.0.3
Author: Ben Pfaff

6.4.5 Aside: Initial Black Insertion |

The traditional algorithm for insertion in an RB tree colors new nodes red. This is a good choice, because it often means that no rebalancing is necessary, but it is not the only possible choice. This section implements an alternate algorithm for insertion into an RB tree that colors new nodes black.

The outline is the same as for initial-red insertion. We change the newly inserted node from red to black and replace the rebalancing algorithm:

    212. <RB item insertion function, initial black 212> =
    void **
    rb_probe (struct rb_table *tree, void *item)
    {
      <rb_probe() local variables 200>

      <Step 1: Search RB tree for insertion point 201>
      <Step 2: Insert RB node; RB_RED => RB_BLACK 202>
      <Step 3: Rebalance after initial-black RB insertion 213>

      return &n->rb_data;
    }

The remaining task is to devise the rebalancing algorithm. Rebalancing is always necessary, unless the tree was empty before insertion, because insertion of a black node into a nonempty tree always violates rule 2. Thus, our invariant is that we have a rule 2 violation to fix.

More specifically, the invariant, as implemented, is that at the top of each trip through the loop, stack pa[] contains the chain of ancestors of a node that is the black root of a subtree whose black-height is 1 more than it should be. We give that node the name q. There is one easy rebalancing special case: if node q has a black parent, we can just recolor q as red, and we're done. Here's the loop:

    213. <Step 3: Rebalance after initial-black RB insertion 213> =
    while (k >= 2)
      {
        struct rb_node *q = pa[k - 1]->rb_link[da[k - 1]];

        if (pa[k - 1]->rb_color == RB_BLACK)
          {
            q->rb_color = RB_RED;
            break;
          }

        if (da[k - 2] == 0)
          {
            <Left-side rebalancing after initial-black RB insertion 214>
          }
        else
          {
            <Right-side rebalancing after initial-black RB insertion 218>
          }
      }

This code is included in 212.

Consider rebalancing where insertion was on the left side of q's grandparent. We know that q is black and its parent pa[k - 1] is red. Then, we can divide rebalancing into three cases, described below in detail. (For additional insight, compare these cases to the corresponding cases for initial-red insertion.)

    214. <Left-side rebalancing after initial-black RB insertion 214> =
    struct rb_node *y = pa[k - 2]->rb_link[1];

    if (y != NULL && y->rb_color == RB_RED)
      {
        <Case 1 in left-side initial-black RB insertion rebalancing 215>
      }
    else
      {
        struct rb_node *x;

        if (da[k - 1] == 0)
          y = pa[k - 1];
        else
          {
            <Case 3 in left-side initial-black RB insertion rebalancing 217>
          }

        <Case 2 in left-side initial-black RB insertion rebalancing 216>
      }

This code is included in 213.

##### Case 1: q's uncle is red

If q has an red “uncle” y, then we recolor q red and pa[k - 1] and y black. This fixes the immediate problem, making the black-height of q equal to its sibling's, but increases the black-height of pa[k - 2], so we must repeat the rebalancing process farther up the tree:

    215. <Case 1 in left-side initial-black RB insertion rebalancing 215> =
    pa[k - 1]->rb_color = y->rb_color = RB_BLACK;
    q->rb_color = RB_RED;
    k -= 2;

This code is included in 214 and 219.

##### Case 2: q is the left child of pa[k - 1]

If q is a left child, then call q's parent y and its grandparent x, rotate right at x, and recolor q, y, and x. The effect is that the black-heights of all three subtrees is the same as before q was inserted, so we're done, and break out of the loop.

    216. <Case 2 in left-side initial-black RB insertion rebalancing 216> =
    x = pa[k - 2];
    x->rb_color = q->rb_color = RB_RED;
    y->rb_color = RB_BLACK;

    x->rb_link[0] = y->rb_link[1];
    y->rb_link[1] = x;
    pa[k - 3]->rb_link[da[k - 3]] = y;
    break;

This code is included in 214.

##### Case 3: q is the right child of pa[k - 1]

If q is a right child, then we rotate left at its parent, which we here call x. The result is in the form for application of case 2, so after the rotation, we relabel the nodes to be consistent with that case.

    217. <Case 3 in left-side initial-black RB insertion rebalancing 217> =
    x = pa[k - 1];
    y = pa[k - 2]->rb_link[0] = q;
    x->rb_link[1] = y->rb_link[0];
    q = y->rb_link[0] = x;

This code is included in 214.

##### 6.4.5.1 Symmetric Case

    218. <Right-side rebalancing after initial-black RB insertion 218> =
    struct rb_node *y = pa[k - 2]->rb_link[0];

    if (y != NULL && y->rb_color == RB_RED)
      {
        <Case 1 in right-side initial-black RB insertion rebalancing 219>
      }
    else
      {
        struct rb_node *x;

        if (da[k - 1] == 1)
          y = pa[k - 1];
        else
          {
            <Case 3 in right-side initial-black RB insertion rebalancing 221>
          }

        <Case 2 in right-side initial-black RB insertion rebalancing 220>
      }

This code is included in 213.

    219. <Case 1 in right-side initial-black RB insertion rebalancing 219> =
    <Case 1 in left-side initial-black RB insertion rebalancing 215>

This code is included in 218.

    220. <Case 2 in right-side initial-black RB insertion rebalancing 220> =
    x = pa[k - 2];
    x->rb_color = q->rb_color = RB_RED;
    y->rb_color = RB_BLACK;

    x->rb_link[1] = y->rb_link[0];
    y->rb_link[0] = x;
    pa[k - 3]->rb_link[da[k - 3]] = y;
    break;

This code is included in 218.

    221. <Case 3 in right-side initial-black RB insertion rebalancing 221> =
    x = pa[k - 1];
    y = pa[k - 2]->rb_link[1] = q;
    x->rb_link[0] = y->rb_link[1];
    q = y->rb_link[1] = x;

This code is included in 218.