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,
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