Difference between revisions of "99 questions/Solutions/61A"

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

Revision as of 13:02, 1 August 2012

Collect the leaves of a binary tree in a list

A leaf is a node with no successors. Write a predicate leaves/2 to collect them in a list.

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

Alternative solution only using cons:

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