Let’s implement a binary tree in Haskell.

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data Tree a = Leaf a
| Node (Tree a) (Tree a)
deriving (Show, Eq)

The above defines a `Tree`

that takes a value of type `a`

. Because we have not explicitly specified any datatype (such as an `Int`

or a `[Char]`

, for instance) and instead defined it in terms of `a`

which can stand for any type, this tree is capable of handling values of any datatype.

It has 2 constructors:

- Constructor for a tree with just a
`Leaf`

. - Constructor for a tree with two subtrees, each of which maybe
- another tree (with further subtrees), or
- a leaf.

We’re deriving `Show`

(line 3) so that we can pretty-print our binary-tree.

It’s evident that defining a tree in Haskell is way more easier than doing the same in Java, C++, or even Python.

Let’s go ahead and create elements using this tree. Save the code above into a file (say `my_tree.hs`

), and load it using the ghci interpreter.

```
$ ghci my_tree.hs
GHCi, version 8.0.1: http://www.haskell.org/ghc/ :? for help
[1 of 1] Compiling Main ( my_tree.hs, interpreted )
Ok, modules loaded: Main.
```

So far so good. Let’s create two trees (using the `Leaf`

value constructor) and see their type.

```
*Main> a = Leaf "cat"
*Main> b = Leaf "rabbit"
*Main> :type a
a :: Tree [Char]
*Main> :type b
b :: Tree [Char]
```

Both `a`

and `b`

are infact `Tree`

that stores `[Char]`

(a.k.a `String`

).

Let’s create a new `Node`

that contains `a`

and `b`

defined above as subtrees.

```
*Main> n = Node a b
*Main> :t n
n :: Tree [Char]
*Main> n
Node (Leaf "cat") (Leaf "rabbit")
```

Creating a `Node`

was simple – we passed in leaves `a`

and `b`

we created earlier into the `Node`

constructor.

The last line in the snippet displays the contents of the tree `n`

– it has two subtrees, each of which is a `Leaf`

. We were able to display the contents in this way because we had derived the `Show`

typeclass.

Note that `n`

is a tree that stores data of type `[Char]`

because all its subtrees store data of type `[Char]`

. If we had instead tried to create a tree with different datatypes, ghci would complain, as the following example demonstrates:

```
*Main> p_num = Leaf 20
*Main> q_str = Leaf "Linda"
*Main> :t p_num
p_num :: Num a => Tree a
*Main> :t q_str
q_str :: Tree [Char]
*Main> r_bad = Node p_num q_str
<interactive>:15:14: error:
• No instance for (Num [Char]) arising from a use of ‘p_num’
• In the first argument of ‘Node’, namely ‘p_num’
In the expression: Node p_num q_str
In an equation for ‘r_bad’: r_bad = Node p_num q_str
```

The problem was that `p_num`

and `q_str`

stores values of different types, and we can’t create a `Tree`

that combines subtrees which store values of two different types.

### Tree Map

Let’s create a `treeMap`

function that maps over a tree, converting values in the leaves to another value.

`treeMap`

should take as arguments:

- A function that converts a value of type
`a`

(ie., a value stored in a`Leaf`

or a`Node`

) to a value of type`b`

;`a`

and`b`

need not necessarily be the same. - A
`Tree`

to work on.

Add the following code to `my_tree.hs`

we were working on earlier.

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treeMap :: (a -> b) -> (Tree a) -> (Tree b)
treeMap f (Leaf a) = Leaf (f a)
treeMap f (Node t1 t2) = Node (treeMap f t1) (treeMap f t2)

Line #2 handles the case when the `Tree`

is a `Leaf`

, and line #3 handles the case when the `Tree`

is a `Node`

.

First up, reload `my_tree.hs`

into the interpreter.

```
*Main> :r
[1 of 1] Compiling Main ( my_tree.hs, interpreted )
Ok, modules loaded: Main.
```

Let’s add some more nodes to the tree we created earlier.

```
*Main> m = Leaf "horse"
*Main> p = Node m n
*Main> p
Node (Leaf "horse") (Node (Leaf "cat") (Leaf "rabbit"))
```

Now, let’s run `treeMap`

on the `p`

to get a new tree with the length of each of the leaves.

```
*Main> treeMap length p
Node (Leaf 5) (Node (Leaf 3) (Leaf 6))
```

Leaves of the new tree contains the string length of values in the original tree `p`

. The structure of the tree remains unchanged.

Since the leaves of the resultant tree stores `Int`

(because `length`

returns values of type `Int`

), we will see this change reflected when we examine its type.

```
*Main> :t (treeMap length p)
(treeMap length p) :: Tree Int
```

### Tree Fold

Let’s now implement a function that folds a tree into a single value.

`treeFold`

should take as arguments:

- Function that accepts a value of type
`a`

and returns a value of type`b`

, to operate on`Leaf`

.

`(a -> b)`

- Function that accepts two values of type
`b`

, and returns a third value of type`b`

. Operates on the values the first function returns (after processing`Leaf`

).

`(b -> b -> b)`

- A
`Tree`

to work on.

Finally, `treeFold`

would return a value of type `b`

.

Append the following code to `my_tree.hs`

:

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treeFold :: (b -> b -> b) -> (a -> b) -> (Tree a) -> b
treeFold fnode fleaf (Leaf n) = fleaf n
treeFold fnode fleaf (Node t1 t2) = fnode (treeFold fnode fleaf t1) (treeFold fnode fleaf t2)

Reload `my_tree.hs`

into the interpreter.

```
*Main> :r
[1 of 1] Compiling Main ( my_tree.hs, interpreted )
Ok, modules loaded: Main.
```

Now, we’ll use `treeFold`

to add the lengths of all elements stored in the leaves of tree `p`

, which we defined earlier.

```
*Main> p
Node (Leaf "horse") (Node (Leaf "cat") (Leaf "rabbit"))
*Main> treeFold (+) length p
14
```

Simple as that!

See the full `my_tree.hs`

below: