# 99 questions/Solutions/62B

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< 99 questions | Solutions(Difference between revisions)

(rewrite a little shorter and (imho) clearer) |
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<haskell> | <haskell> | ||

− | + | atLevel :: Tree a -> Int -> [a] | |

− | + | atLevel Empty _ = [] | |

− | + | atLevel (Branch v l r) n | |

| n == 1 = [v] | | n == 1 = [v] | ||

| n > 1 = atlevel l (n-1) ++ atlevel r (n-1) | | n > 1 = atlevel l (n-1) ++ atlevel r (n-1) | ||

Line 19: | Line 19: | ||

levels (Branch a l r) = [a] : zipWith (++) (levels l) (levels r) | levels (Branch a l r) = [a] : zipWith (++) (levels l) (levels r) | ||

− | + | atLevel :: Tree a -> Int -> [a] | |

− | + | atLevel t n = levels t !! (n-1) | |

</haskell> | </haskell> |

## Latest revision as of 08:49, 2 December 2010

Collect the nodes at a given level in a list

A node of a binary tree is at level N if the path from the root to the node has length N-1. The root node is at level 1. Write a predicate atlevel/3 to collect all nodes at a given level in a list.

atLevel :: Tree a -> Int -> [a] atLevel Empty _ = [] atLevel (Branch v l r) n | n == 1 = [v] | n > 1 = atlevel l (n-1) ++ atlevel r (n-1) | otherwise = []

Another possibility is to decompose the problem:

levels :: Tree a -> [[a]] levels Empty = repeat [] levels (Branch a l r) = [a] : zipWith (++) (levels l) (levels r) atLevel :: Tree a -> Int -> [a] atLevel t n = levels t !! (n-1)