[Haskell-beginners] Hutton 2016 ex8.3a

[trent shipley]

data Tree a = Leaf a | Node (Tree a) (Tree a)

How could I modify this definition so that a Node could be:

Node (Tree a) Null/Nothing/[] -- basically a zilch that would match
anything or nothing, because it would make programming simple.
Alternatively:

Node (Tree a)

-- Would I be looking at,

data Tree a = Leaf a | Node (Tree a)  Maybe (Tree a)

-- Or possibly the best I can do is,

data Tree a = Leaf a | Node (Tree a) (Tree a) | Node (Tree a)









On Mon, Sep 3, 2018 at 3:06 AM Rishi Rajasekaran <
rajasekaran.rishi at gmail.com> wrote:

> Hi Trent
>
> For executing your approach, you can do the following:
> 1. Compute the number of leaves in the left subtree first.
> 2. Pass that computed value into leaves' call for the right subtree
>
> Regards
>
> Rishi
>
>
> On Mon, 3 Sep, 2018, 2:36 PM trent shipley, <trent.shipley at gmail.com>
> wrote:
>
>> Given
>> data Tree a = Leaf a | Node (Tree a) (Tree a)
>>
>> Write a leaf counter.
>>
>> Hutton suggests:
>>
>> leaves :: Tree a -> Int
>> leaves (Leaf _) = 1
>> leaves (Node l r) = leaves l + leaves r
>>
>> I tried:
>>
>> leavesTrent :: Tree a -> Int
>> leavesTrent = leaves' 0
>>          where
>>            leaves' n (Leaf a) = n + 1
>>            leaves' n (Node l r) = (leaves' n l), (leaves' n r)
>>
>> The idea is:
>>
>> If it is a leaf, add one to the accumulator.  (Following Hutton's
>> explanation of how sum works if defined with foldl.)   If it is a tree,
>> proceed down the left subtree recursively, until you get to a leaf, then
>> roll up to the right subtree.  The problem (among the problems) is that I
>> don't know how to tell the compiler to do all lefts, before backing up to
>> go right.  I only know how to do that using a real operator like "+" or foo
>> (l, r).
>>
>> Is that kind of no-op recursive branching possible?
>>
>> Trent.
>>
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