CS 312 Lecture ?:
AVL Trees

Binary search trees

A binary search tree is one in which every node n satisfies the binary search tree invariant: its left child and all the nodes below it have values (or keys) less than that of n. Similarly, the right child node and all nodes below it have values greater than that of n.

The code for a binary search tree looks like the following. First, to check for an element or to add a new element, we simply walk down the tree.

(* contains(t,x) is whether x is in the tree *)
fun contains(t: tree, x: value): bool =
  case t of
    Empty => false
  | Node{value, left, right} =>
      (case compare(x, value) of
         EQUAL => true
       | LESS => contains(left, x)
       | GREATER => contains(right, x))

(* add(t,x) is a BST with the same values as t, plus x *)
fun add(t: tree, x: value): tree = let
  fun balance(t: tree): tree = t (* what to write here? *)
  case t of
    Empty => Node{value=x, left=Empty, right=Empty}
  | Node {value, left, right} =>
    (case compare(x, value) of
      EQUAL => Node{value=x, left=left, right=right}
    | LESS => Node{value=value, left=add(left, x), right=right}
    | GREATER => Node{value=value, left=left, right=add(right,x) } )

When a tree satisfies the BST invariant, an in-order traversal of the tree nodes will visit the nodes in ascending order of their contained values. So it's easy to fold over all tree nodes in order.

Removing elements is a little trickier. If a node is a leaf, it can be removed. If it has one child, it can be replaced with its child. If it has two children, it can be replaced with either its immediate successor (or predecessor), which requires searching in the tree.

(* Returns: a tree just like t except that the node containing x is removed.
 * Checks: x is in the tree. *)
fun remove(t: tree, x: value): tree =
    (* Returns: a tree in which the successor of the root is removed,
     * along with the value of that successor.
     * Checks: the root has a successor. *)
    fun removeSuccessor(t: tree): tree*value =
      case t
        of Empty => raise Fail "impossible"
         | Node {value, left=Empty, right} => (r, v)
         | Node {value, left, right} => let val (l, v) = removeSuccessor(l)
                                        in (Node {value, l, right}, v) end
    case t
      of Empty => raise Fail "value not in the tree"
       | Node {value, left, right) =>
           case Int.compare(x, value)
             of LESS => Node {value, remove(l, x), right)
              | GREATER => Node {value, left, remove(r, x)}
              | EQUAL => case (left, right)
                           of (_, Empty) => l
                            | (Empty, _) => r
                            | _ => let val (r, v) = removeSuccessor(r) in
				     Node {v, left, r}

The time required to find a node in a BST, or to remove a node from a BST, is O(h), where h is the height of the tree: the length of the longest path from the root node to any leaf. If a tree is perfectly balanced, so that all leaf nodes are at the same depth, then h is O(log n). This makes binary search trees an attractive data structure, especially for implementing ordered sets and maps.

The problem with BST's is that they are not necessarily balanced. In fact, if nodes are added to a BST in increasing order, the resulting BST will be essentially a linked list. The solution to this problem is to make sure that the BST is balanced. Making the BST perfectly balanced at every step is too expensive, but if we are interested in asymptotic complexity, we merely need the height h to be proportional to O(log n). We will say that the BST is balanced in this case.

There are many ways to keep binary search trees balanced. Some of the more popular methods are red-black trees, AVL trees, B-trees, and splay trees. But there are many more, including 2-3 trees, 2-3-4 trees, AA trees, and treaps. Each kind of binary search tree works by strengthening the representation invariant so that the tree must be approximately balanced.

AVL trees

AVL trees were invented by Adelson-Velskii and Landis in 1962. An AVL tree is a balanced binary search tree where every node in the tree satisfies the following invariant: the height difference between its left and right children is at most 1. Hence, all sub-trees of an AVL tree are themselves AVL. The height difference between children is referred to as the balance factor of the node.

Let's see why the AVL invariant means that the tree is balanced. Suppose we want to make it as unbalanced as possible. Then for a given height h, we want to find the AVL tree with as few nodes as possible. Let N(h) be the minimum number of nodes in a tree of height h. If we think about it, we can see that N(0)=1, N(1)=2, N(2)=4, and in general,

N(h) = 1 + N(h−1) + N(h−2)

because a tree with height h must have at least one child with height h−1, and to make the tree as small as possible, we make the other child have height h−2.

Now we can show that there is a minimum size to an AVL tree of height h. We do this by showing that N(h) has a lower bound; that it is Ω(kh) for suitable k.

We use the substitution method, replacing N(h) with ckh on both sides of the recurrence. We need to find c, h0 such that for all h greater than h0,

c kh ≤ 1 + ckh−1 + ckh−2

Dividing by ckh−2, we see this is true if

k2k-h/c + k + 1

Because the term k-h/c becomes small for large h, this inequality will hold as long as k is less than the solution to the equation:

k2 = k + 1

which is the golden ratio, φ = 1.618... . Therefore, n is Ω(φh), and conversely h is O(logφn) = O(lg n). Therefore an AVL tree is balanced.

You may know that the golden ratio is connected to the Fibonacci series. If you look more closely at the function N(h), you'll notice that in fact N(h) = F(h+2) - 1, where F(n) gives the nth Fibonacci number.

Balancing AVL trees

Here is the code for AVL trees. The key piece of technology is the balance function, which rebalances an AVL tree. All the operations such as add and remove can then use balance to restore the AVL invariant.

type height = int
datatype avltree = Empty | Node of height * value * avltree * avltree

(* Rep Invariant:
 * For each node Node(h, v, l, r):
 * (1) BST invariant: v is greater than all values in l,
 *                    and less than all values in r.
 * (2) h is the height of the node.
 * (3) Each node is balanced, i.e., abs(l.h - r.h) <= 1

fun height(Empty) = 0
  | height(Node(h,_,_,_)) = h

fun bal_factor(Empty) = 0
  | bal_factor(Node(_,_,l,r)) = (height l) - (height r)

fun node(v: value, l: avltree, r: avltree): avltree =
  Node(1+Int.max(height l, height r), v, l, r)

fun rotate_left(t: avltree): avltree =
  case t 
    of Node(_, x, a, Node(_, y, b, c)) => node(y, node(x, a, b), c)
     | _ => t

fun rotate_right(t: avltree): avltree =
  case t
    of Node(_, x, Node(_, y, a, b), c) => node(y, a, node(x, b, c))
     | _ => t

(* Returns: an AVL tree containing the same values as n.
 * Requires: The children of n satisfy the AVL invariant, and
 *           their heights differ by at most 2. *)
fun balance(n as Node(h, v, l, r): avltree): avltree =
  case (bal_factor n, bal_factor l, bal_factor r) 
    of ( 2, ~1, _) => rotate_right(node(v, rotate_left l, r))
     | ( 2, _, _)  => rotate_right(n)
     | (~2, _, 1)  => rotate_left (node(v, l, rotate_right r))
     | (~2, _, _)  => rotate_left (n)
     | _ => n

fun add (t: avltree, n:int): avltree =
  case t
    of Empty => node(n, Empty, Empty)
     | Node(h, v, l, r) =>
         case Int.compare (n, v)
           of EQUAL => t
            | LESS => balance(node(v, add(l, n), r))
            | GREATER => balance(node(v, l, add(r, n)))

fun remove(t: avltree, n: int): avltree =
    fun removeSuccessor(t: avltree): avltree*int =
      case t
        of Empty => raise Fail "impossible"
         | Node(_, v, Empty, r) => (r, v)
         | Node(_, v, l, r) => let val (l', v') = removeSuccessor(l)
                               in (balance(node(v, l', r)), v') end
    case t
      of Empty => raise Fail "value not in the tree"
       | Node (_, v, l, r) =>
           case Int.compare(n, v)
             of LESS => balance(node(v, remove(l, n), r))
              | GREATER => balance(node(v, l, remove(r, n)))
              | EQUAL => case (l, r)
                           of (_, Empty) => l
                            | (Empty, _) => r
                            | _ => let val (r', v') = removeSuccessor(r)
                                   in balance(node(v', l, r')) end

The balance function works by doing tree rotations. This is a reorganizing of the tree in which the parent-child relationships between nodes are changed in a local way, usually to restore a global invariant. There are two basic tree rotations, left rotations and right rotations, which are symmetrical. A left rotation works as follows, moving the root node to the left:

     x                       y
   /   \                   /   \ 
  +     y                 x     +
 /a\   / \       ===>    / \   /c\
 ---  +   +             +   +  ---
     /b\ /c\           /a\ /b\
     --- ---           --- ---

A right rotation is just the inverse transformation. The important property is that tree rotations preserve the BST invariant, because the left-to-right ordering of all nodes remains unchanged: x < a < b < y < c Therefore tree rotations can be used to reestablish other invariants such as the AVL invariant.

The balance function is invoked on a node t that is possibly unbalanced. We assume that whatever operation has been performed on the tree below this node, it has changed the height of nodes by at most one, and therefore the child subtrees of t have a height difference of at most 2. We also assume the subtrees satisfy the AVL invariant themselves. If the height difference (balance factor) is 1 or 0, then balance doesn't need to do anything. Suppose the balance factor is 2 (the case where it is −2 is symmetrical). Then the tree t looks something like this:

        /     \  
       +       + 
      / \     / \
     /   \   / h \
    / h+2 \  -----
   /       \

How we fix this problem depends on what the left subtree looks like. There are two cases to consider:

        Case 1                    Case 2
           y                         y
        /     \                   /     \    
       x       +                 x       z   
     /   \    /h\              /   \    /h\  
    +     +   ---             +     +   ---  
   / \   / \   c             /h\   / \   c   
  /h+1\ /h+1\                ---  /h+1\      
  ----- -----                 a   -----      
    a     b (may be h)              b         
In Case 1, we can do a right rotation to pull the subtrees a and b up:
Case 1:
          y                         x
	/   \                     /   \
       /     \                   /     \
      x       +                 +        y    
    /   \    /h\     ====>     / \     /   \   
   +     +   ---              /h+1\   +     +
  / \   / \   c               -----  / \   /h\
 /h+1\ /h+1\                    a   /h+1\  ---
 ----- -----                        -----   c
   a     b (may be h)                 b

This clearly wouldn't work if the height of subtree a were h, because in that case b's leaves would be two levels lower than a's. That's the job of Case 2, which requires a double rotation:

Case 2:
       z                  y
     /   \              /   \
    /     \            /     \  
   x       +          x        z 
  /  \    /h\ ====>  / \      / \
 +    y   ---       +   +    +   +
/h\  / \   c       /h\ /h\  /h\ /h\
--- +   +          --- ---  --- --- 
 a /h\ /h\          a   b'  b''  c
   --- ---            
    b'  b''          

(Note that one of b' or b'' can actually have height h−1 here, but that doesn't break the AVL invariant). The double rotation preserves the BST ordering because it is equivalent to two rotations. So the ordering remains unchanged: a < x < b' < y < b'' < z < c

More resources

When writing tree algorithms, it's helpful to be able to print out trees on the display. Here is some code for visualizing trees. Call print(tree) to produce some nice output.

For example, it produces output like this:
    / \   
   /   \  
   3   7  
  / \ / \ 
  2 4 6 8 
 /       \
 1       9