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@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, 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.

_______________________________________________

...

Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners

_______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners