3.11. Example: Trees

Trees are a very useful data structure. A binary tree, as you’ll recall from CS 2110, is a node containing a value and two children that are trees. A binary tree can also be an empty tree, which we also use to represent the absence of a child node.

3.11.1. Representation with Tuples

Here is a definition for a binary tree data type:

type 'a tree =
| Leaf
| Node of 'a * 'a tree * 'a tree
type 'a tree = Leaf | Node of 'a * 'a tree * 'a tree

A node carries a data item of type 'a and has a left and right subtree. A leaf is empty. Compare this definition to the definition of a list and notice how similar their structure is:

type 'a tree =                        type 'a mylist =
  | Leaf                                | Nil
  | Node of 'a * 'a tree * 'a tree      | Cons of 'a * 'a mylist

The only essential difference is that Cons carries one sublist, whereas Node carries two subtrees.

Here is code that constructs a small tree:

(* the code below constructs this tree:
         4
       /   \
      2     5
     / \   / \
    1   3 6   7
*)
let t =
  Node(4,
    Node(2,
      Node(1, Leaf, Leaf),
      Node(3, Leaf, Leaf)
    ),
    Node(5,
      Node(6, Leaf, Leaf),
      Node(7, Leaf, Leaf)
    )
  )
val t : int tree =
  Node (4, Node (2, Node (1, Leaf, Leaf), Node (3, Leaf, Leaf)),
   Node (5, Node (6, Leaf, Leaf), Node (7, Leaf, Leaf)))

The size of a tree is the number of nodes in it (that is, Nodes, not Leafs). For example, the size of tree t above is 7. Here is a function function size : 'a tree -> int that returns the number of nodes in a tree:

let rec size = function
  | Leaf -> 0
  | Node (_, l, r) -> 1 + size l + size r

3.11.2. Representation with Records

Next, let’s revise our tree type to use use a record type to represent a tree node. In OCaml we have to define two mutually recursive types, one to represent a tree node, and one to represent a (possibly empty) tree:

type 'a tree =
  | Leaf
  | Node of 'a node

and 'a node = {
  value: 'a;
  left: 'a tree;
  right: 'a tree
}
type 'a tree = Leaf | Node of 'a node
and 'a node = { value : 'a; left : 'a tree; right : 'a tree; }

Here’s an example tree:

(* represents
      2
     / \
    1   3  *)
let t =
  Node {
    value = 2;
    left = Node {value = 1; left = Leaf; right = Leaf};
    right = Node {value = 3; left = Leaf; right = Leaf}
  }
val t : int tree =
  Node
   {value = 2; left = Node {value = 1; left = Leaf; right = Leaf};
    right = Node {value = 3; left = Leaf; right = Leaf}}

We can use pattern matching to write the usual algorithms for recursively traversing trees. For example, here is a recursive search over the tree:

(** [mem x t] is whether [x] is a value at some node in tree [t]. *)
let rec mem x = function
  | Leaf -> false
  | Node {value; left; right} -> value = x || mem x left || mem x right
val mem : 'a -> 'a tree -> bool = <fun>

The function name mem is short for “member”; the standard library often uses a function of this name to implement a search through a collection data structure to determine whether some element is a member of that collection.

Here’s a function that computes the preorder traversal of a tree, in which each node is visited before any of its children, by constructing a list in which the values occur in the order in which they would be visited:

let rec preorder = function
  | Leaf -> []
  | Node {value; left; right} -> [value] @ preorder left @ preorder right
val preorder : 'a tree -> 'a list = <fun>
preorder t
- : int list = [2; 1; 3]

Although the algorithm is beautifully clear from the code above, it takes quadratic time on unbalanced trees because of the @ operator. That problem can be solved by introducing an extra argument acc to accumulate the values at each node, though at the expense of making the code less clear:

let preorder_lin t =
  let rec pre_acc acc = function
    | Leaf -> acc
    | Node {value; left; right} -> value :: (pre_acc (pre_acc acc right) left)
  in pre_acc [] t
val preorder_lin : 'a tree -> 'a list = <fun>

The version above uses exactly one :: operation per Node in the tree, making it linear time.