An alternative layout method is depicted in the illustration below:
Find out the rules and write the corresponding function. Hint: On a given level, the horizontal distance between neighboring nodes is constant.
Use the same conventions as in problem P64 and test your function in an appropriate way.
layout :: Tree a -> Tree (a, Pos) layout t = layoutAux x1 1 sep1 t where d = depth t ld = leftdepth t x1 = 2^(d-1) - 2^(d-ld) + 1 sep1 = 2^(d-2) layoutAux x y sep Empty = Empty layoutAux x y sep (Branch a l r) = Branch (a, (x,y)) (layoutAux (x-sep) (y+1) (sep `div` 2) l) (layoutAux (x+sep) (y+1) (sep `div` 2) r) depth :: Tree a -> Int depth Empty = 0 depth (Branch a l r) = max (depth l) (depth r) + 1 leftdepth :: Tree a -> Int leftdepth Empty = 0 leftdepth (Branch a l r) = leftdepth l + 1
The auxiliary function is passed the x- and y-coordinates for the root of the subtree, the horizontal separation between the root and its child nodes, and the subtree itself. It returns the subtree annotated with positions.
Another solution is presented below. The root is placed at (0, 0). (I assume this solution uses more memory, since we pass the list of children into the recursion for each subtree. Can somebody verify this?)
tree65 t = helper [t] 0 0 t where helper ss x y Empty = Empty helper ss x y (Branch _ t0 t1) = let r = foldr1 max $ map radius ss ss' = concatMap children ss in Branch (x, y) (helper ss' (x - r) (y + 1) t0) (helper ss' (x + r) (y + 1) t1) children (Branch _ t0 t1) = [t0, t1] children Empty =  radius Empty = 1 radius (Branch _ t0 t1) = let w0 = 2 * (radius t0) - 1 w1 = 2 * (radius t1) - 1 in max w0 w1 + 1