99 questions/Solutions/62: Difference between revisions

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internals (Branch a Empty Empty) = []
internals (Branch a Empty Empty) = []
internals (Branch a left right ) = a : internals left ++ internals right
internals (Branch a left right ) = a : internals left ++ internals right
</haskell>
Alternative solution only using cons:
<haskell>
internals t = internals' t []
    where internals'  Empty                xs = xs
          internals' (Branch x Empty Empty) xs = xs
          internals' (Branch x l r)        xs = (x :) $ internals' l $ internals' r xs
</haskell>
</haskell>

Revision as of 13:06, 1 August 2012

Collect the internal nodes of a binary tree in a list

An internal node of a binary tree has either one or two non-empty successors. Write a predicate internals/2 to collect them in a list.

internals :: Tree a -> [a]
internals Empty                  = []
internals (Branch a Empty Empty) = []
internals (Branch a left right ) = a : internals left ++ internals right

Alternative solution only using cons:

internals t = internals' t []
    where internals'  Empty                 xs = xs
          internals' (Branch x Empty Empty) xs = xs
          internals' (Branch x l r)         xs = (x :) $ internals' l $ internals' r xs