# 99 questions/Solutions/59

### From HaskellWiki

(**) Construct height-balanced binary trees

In a height-balanced binary tree, the following property holds for every node: The height of its left subtree and the height of its right subtree are almost equal, which means their difference is not greater than one.

hbalTree x = map fst . hbalTree' where hbalTree' 0 = [(Empty, 0)] hbalTree' 1 = [(Branch x Empty Empty, 1)] hbalTree' n = let t = hbalTree' (n-2) ++ hbalTree' (n-1) in [(Branch x lb rb, h) | (lb,lh) <- t, (rb,rh) <- t , let h = 1 + max lh rh, h == n]

Alternative solution:

hbaltree :: a -> Int -> [Tree a] hbaltree x 0 = [Empty] hbaltree x 1 = [Branch x Empty Empty] hbaltree x h = [Branch x l r | (hl, hr) <- [(h-2, h-1), (h-1, h-1), (h-1, h-2)], l <- hbaltree x hl, r <- hbaltree x hr]

If we want to avoid recomputing lists of trees (at the cost of extra space), we can use a similar structure to the common method for computation of all the Fibonacci numbers:

hbaltree :: a -> Int -> [Tree a] hbaltree x h = trees !! h where trees = [Empty] : [Branch x Empty Empty] : zipWith combine (tail trees) trees combine ts shortts = [Branch x l r | (ls, rs) <- [(shortts, ts), (ts, ts), (ts, shortts)], l <- ls, r <- rs]