Red-Black Trees

Red-black trees are relatively simple balanced binary tree data structure. The idea is to strengthen the representation invariant so a tree has height logarithmic in the number of nodes nn. To help enforce the invariant, we color each node of the tree either red or black. Where it matters, we consider the color of an empty tree to be black.

type color = Red | Black
type 'a rbtree =
  Node of color * 'a * 'a rbtree * 'a rbtree | Leaf

Here are the new conditions we add to the binary search tree representation invariant:

  1. There are no two adjacent red nodes along any path.
  2. Every path from the root to a leaf has the same number of black nodes. This number is called the black height (BH) of the tree.

If a tree satisfies these two conditions, it must also be the case that every subtree of the tree also satisfies the conditions. If a subtree violated either of the conditions, the whole tree would also.

Additionally, by convention the root of the tree is colored black. This does not violate the invariants, but it also is not required by them.

With these invariants, the longest possible path from the root to an empty node would alternately contain red and black nodes; therefore it is at most twice as long as the shortest possible path, which only contains black nodes. The longest path cannot have a length greater than twice the length of the paths in a perfect binary tree, which is O(logn)O(\log n). Therefore, the tree has height O(logn)O(\log n) and the operations are all asymptotically logarithmic in the number of nodes.

How do we check for membership in red-black trees? Exactly the same way as for general binary trees.

let rec mem x = function
  | Leaf -> false
  | Node (_, y, l, r) ->
    if x < y then mem x l
    else if x > y then mem x r
    else true

More interesting is the insert operation. As with standard binary trees, we add a node by replacing the leaf found by the search procedure. We also color the new node red to ensure that invariant 2 is preserved. However, this may destroy invariant 1 by producing two adjacent red nodes. In order to restore the invariant, we consider not only the new red node and its red parent, but also its (black) grandparent. The next figure shows the four possible cases that can arise.

           1             2             3             4

           Bz            Bz            Bx            Bx
          / \           / \           / \           / \
         Ry  d         Rx  d         a   Rz        a   Ry
        /  \          / \               /  \          /  \
      Rx   c         a   Ry            Ry   d        b    Rz
     /  \               /  \          / \                /  \
    a    b             b    c        b   c              c    d

Notice that in each of these trees, the values of the nodes in a, b, c, and d must have the same relative ordering with respect to x, y, and z: a < x < b < y < c < z < d. Therefore, we can transform the tree to restore the invariant locally by replacing any of the above four cases with:

        /  \
      Bx    Bz
     / \   / \
    a   b c   d

This balance function can be written simply and concisely using pattern matching, where each of the four input cases is mapped to the same output case. In addition, there is the case where the tree is left unchanged locally.

let balance = function
  | Black, z, Node (Red, y, Node (Red, x, a, b), c), d
  | Black, z, Node (Red, x, a, Node (Red, y, b, c)), d
  | Black, x, a, Node (Red, z, Node (Red, y, b, c), d)
  | Black, x, a, Node (Red, y, b, Node (Red, z, c, d)) ->
      Node (Red, y, Node (Black, x, a, b), Node (Black, z, c, d))
  | a, b, c, d -> Node (a, b, c, d)

This balancing transformation possibly breaks invariant 1 one level up in the tree, but it can be restored again at that level in the same way, and so on up the tree. In the worst case, the process cascades all the way up to the root, resulting in two adjacent red nodes, one of them the root. But if this happens, we can just recolor the root black, which increases the BH by one. The amount of work is O(logn)O(\log n). The insert code using balance is as follows:

let insert x s =
  let rec ins = function
    | Leaf -> Node (Red, x, Leaf, Leaf)
    | Node (color, y, a, b) as s ->
      if x < y then balance (color, y, ins a, b)
      else if x > y then balance (color, y, a, ins b)
      else s in
  match ins s with
    | Node (_, y, a, b) -> Node (Black, y, a, b)
    | Leaf -> (* guaranteed to be nonempty *)
        failwith "RBT insert failed with ins returning leaf"

Removing an element from a red-black tree works analogously. We start with BST element removal and then do rebalancing. When an interior (nonleaf) node is removed, we simply splice it out if it has fewer than two nonleaf children; if it has two nonleaf children, we find the next value in the tree, which must be found inside its right child.

Balancing the trees during removal from red-black tree requires considering more cases. Deleting a black element from the tree creates the possibility that some path in the tree has too few black nodes, breaking the black-height invariant 2. The solution is to consider that path to contain a "doubly-black" node. A series of tree rotations can then eliminate the doubly-black node by propagating the "blackness" up until a red node can be converted to a black node, or until the root is reached and it can be changed from doubly-black to black without breaking the invariant.

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