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 [hidden email] http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners |
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, <[hidden email]> wrote:
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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 <[hidden email]> wrote:
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