Haskell: Get path from every leaf node to root in a Bin tree structure

问题

I've got work to do, but i dunno how to do it. I have Bin tree

    1
   /  \
  2    3
 / \  / \
4  5  6  7

And I need to find the way from root to Node with the coord [i, j]. For example: (2, 2) -> [1, 3, 6]

fromRoot :: Int -> Int -> Tree a -> [a]

I wrote some function for Index and BinTree, but how to make main function i dont know.

data Tree a = Node (Tree a) a (Tree a)

index :: Tree a -> Int -> Int -> a
index (Node _ x _ ) 0 _ = x
index (Node l x r) i j  | ((border i)<j) = index r (i-1) (j-(border i)-1)       
                        | otherwise = index l (i-1) j


border :: Int -> Int
border 0 = 0
border 1 = 0
border l = 2*(border (l-1))+1

myBuild :: Int  -> Tree Int
myBuild n = (Node (myBuild (n*2)) n (myBuild (n*2+1)))

回答1:

Since this is homework I won't give a complete solution, but some hints:

• how do you represent an empty tree with your Tree type?
• how do you represent the example tree (or any other finite tree)?

Considering the main function: you don't necessarily need one, a good way to start is running

ghci your_source_file.hs

You can then evaluate parts of your program, e.g.:

fromRoot 2 3 t1 -- if you have a t1 is a tree

Apart from that you could write a main function like this:

test_tree = ...   -- you need to fill in the dots (see questions above)

main :: IO ()
main = do print (fromRoot 2 2 test_tree)

If you need to find some documentation, use http://haskell.org/hoogle/

来源：https://stackoverflow.com/questions/8077802/haskell-get-path-from-every-leaf-node-to-root-in-a-bin-tree-structure