Difference between revisions of "99 questions/Solutions/62"
< 99 questions | Solutions
Jump to navigation
Jump to search
(categorize) |
|||
Line 18: | Line 18: | ||
internals' (Branch x l r) xs = (x :) $ internals' l $ internals' r xs |
internals' (Branch x l r) xs = (x :) $ internals' l $ internals' r xs |
||
</haskell> |
</haskell> |
||
+ | |||
+ | |||
+ | [[Category:Programming exercise spoilers]] |
Latest revision as of 13:39, 25 December 2016
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