CS 3110 Lecture 11
Balanced Binary Trees: Red-Black Trees

Sets and maps are important and useful abstractions. We've seen various ways to implement an abstract data type for sets and maps, since data structures that implement sets can be used to implement maps as well. Today we will look at an implementation of sets that is asymptotically efficient and useful in practice. This implementation is one of several balanced binary tree schemes.

Binary trees have two advantages above the asymptotically more efficient hash table: first, they support nondestructive update with the same asymptotic efficiency. Second, they store their values (or keys, in the case of a map) in order, which makes range queries and in-order iteration possible.

An important property of a search tree is that it can be used to implement an ordered set or ordered map easily: a set (map) that abstractly keeps its elements in sorted order. Although we will not consider such operations today, ordered sets generally provide operations for finding the minimum and maximum elements of the set, for iterating over all the elements between two elements, and for extracting (or iterating over) ordered subsets of the elements between a range:

Binary search trees

A binary tree is easy to define inductively in OCaml. We will use the following definition which represents a node as a triple of a value and two children, and which explicitly represents leaf nodes.
type 'a tree = TNode of 'a * 'a tree * 'a tree | TLeaf

A binary search tree is a binary tree with the following representation invariant: For any node n, every node in the left subtree of n has a value less than that of n, and every node in the right subtree of n has a value more than that of n.

Given such a tree, how do you perform a lookup operation? Start from the root, and at every node, if the value of the node is what you are looking for, you are done; otherwise, recursively look up in the left or right subtree depending on the value stored at the node. In code:

let rec contains x = function
    TLeaf -> false
  | TNode (y, l, r) ->
      if x=y then true else if x < y then contains x l else contains x r

Note the use of the keyword function so that the variable used in the pattern matching need not be named. This is equivalent to (unneccessarily) naming a variable and then using match:

let rec contains x t =
  match t with
      TLeaf -> false
    | TNode (y, l, r) ->
        if x=y then true else if x < y then contains x l else contains x r

Adding an element is similar: you perform a lookup until you find the empty node that should contain the value. This is a nondestructive update, so as the recursion completes, a new tree is constructed that is just like the old one except that it has a new node (if needed):

let rec add x = function
    TLeaf -> TNode (x, TLeaf, TLeaf) (* When get to leaf, put new node there *)
  | TNode (y, l, r) as t -> (* Recursively search for value *)
      if x=y then t
      else if x > y then TNode (y, l, add x r)
      else (* x < y *) TNode (y, add x l, r)

What is the running time of those operations? Since add is just a lookup with an extra constant-time node creation, we focus on the lookup operation. Clearly, the run time of lookup is O(h), where h is the height of the tree. What's the worst-case height of a tree? Clearly, a tree of n nodes all in a single long branch (imagine adding the numbers 1,2,3,4,5,6,7 in order into a binary search tree). So the worst-case running time of lookup is still O(n) (for n the number of nodes in the tree).

What is a good shape for a tree that would allow for fast lookup? A perfect binary tree has the largest number of nodes n for a given height h: n = 2h+1-1. Therefore h = lg(n+1)-1 = O(lg n).

          ^                   50
          |               /        \
          |           25              75
 height=3 |         /    \          /    \
  n=15    |       10     30        60     90
          |      /  \   /  \      /  \   /  \
          V     4   12 27  40    55  65 80  99

If a tree with n nodes is kept balanced, its height is O(lg n), which leads to a lookup operation running in time O(lg n).

How can we keep a tree balanced? It can become unbalanced during element addition or deletion. Most balanced tree schemes involve adding or deleting an element just like in a normal binary search tree, followed by some kind of tree surgery to rebalance the tree. Some examples of balanced binary search tree data structures include

In each of these, we ensure asymptotic complexity of O(lg n) by enforcing a stronger invariant on the data structure than just the binary search tree invariant.

Red-Black Trees

Red-black trees are a fairly simple and very efficient data structure for maintaining a balanced binary tree. The idea is to strengthen the rep invariant so a tree has height logarithmic in n. 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 rep invariant:

  1. No red node has a red parent (alternatively a red node has black children).
  2. Every path from the root to an empty node has the same number of black nodes: the black height of the tree. Call this BH.

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 we will require that the root of the tree be colored black, which note can always be done by simply setting its color to black.

A repOK function for checking this invariant must check the ordering property of regular binary search trees as well as the two properties above. Such code is included in the lec11.ml.

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. If n is the number of nodes in the tree, the longest path cannot have a length greater than twice the length of the paths in a perfect binary tree: 2 lg n, which is O(lg n). Therefore, the tree has height O(lg n) and the operations are all as asymptotically efficient as we could expect.

Another way to see this is to think about just the black nodes in the tree. Suppose we snip all the red nodes out of the trees by connecting black nodes to their closest black descendants. Then we have a tree whose leaves are all at depth BH, and whose branching factor ranges between 2 and 4. Such a tree must contain at least Ω(2BH) nodes, and so must the whole tree when we add the red nodes back in. If N is Ω(2BH), then black height BH is O(lg N). But invariant 1 says that the longest path is at most h = 2BH. So h is O(lg N) too.

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, left, right) ->
      x = y || (x < y && mem x left) || (x > y && mem x right)

More interesting is the insert operation. As with standard binary trees we add a node by replacing the empty node 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 red nodes, one the parent of the other. In order to restore this invariant we consider not only the two red node that has a red parent, but the (black) grandparent. Otherwise, the red-red conflict cannot be fixed while preserving black depth. The next figure shows the four possible cases that may arise of a red node with a red parent and black grandparent:

       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,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 (which can again be fixed at that level). By performing a rebalance of the tree where a new red node is introduced and at all the levels above, we can locally (and incrementally) enforce invariant #1. In the end, we may end up with two red nodes, one of them the root and the other the child of the root; this we can easily correct by coloring the root black. The insert code (using balance) is thus 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
    match ins s with
	Node (_, y, a, b) ->
	  Node (Black, y, a, b)
      | Leaf -> (* guaranteed to be nonempty *)
	  raise (Failure "RBT insert failed with ins returning leaf")

Removing elements

Removing an element from a red-black tree works analogously. We start with BST element removal and then do rebalancing. Here is code to remove elements from a binary tree. The key is that when an interior (nonleaf) node is removed, then we simply splice it out if it has zero or one children; if it has two 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.