Summation of an integer tree (Haskell)

Question

I am trying to make a function that sums up the values ​​of a non binary integer tree.

-- datastructures.hs    
data Tree a = Empty | Node a [Tree a] deriving (Eq, Show)

myNums :: (Num a) => Tree a
myNums = Node 1 [ 
           Node 2 [ 
             Node 4 [Empty], Node 5 [Empty]
           ], 
           Node 3 [
             Node 6 [Empty], Node 7 [Empty], Node 8 [Empty] 
           ]
        ]

addNums :: (Num a) => Tree a -> a
addNums Empty = 0
addNums (Node n [Empty]) = n
addNums (Node n (x:xs)) = n + (addNums x) + (addNums xs)

Ideally, I would like to addNums myNumsbe 36, but this causes an error:

datastructures.hs:20:54:
    Couldn't match expected type ‘Tree a’ with actual type ‘[Tree a]’
    Relevant bindings include
      xs :: [Tree a] (bound at datastructures.hs:20:20)
      x :: Tree a (bound at datastructures.hs:20:18)
      n :: a (bound at datastructures.hs:20:15)
      addNums :: Tree a -> a (bound at datastructures.hs:18:1)
    In the first argument of ‘addNums’, namely ‘xs’
    In the second argument of ‘(+)’, namely ‘(addNums xs)’

How do I counter this, and what are the best practices?

EDIT: Best practices don't seem to be accepted at all Empty! I forgot what []is a valid type instance [Tree a]. Therefore, the best way to implement this is:

data Tree a = Node a [Tree a] deriving (Eq, Show)

addNums :: (Num a) => Tree a -> a
addNums (Node n []) = n
addNums (Node n (x:xs)) = n + (addNums x) + addNums (Node 0 xs)

Answer 1

addNums. , Empty . - :

addNums :: (Num a) => Tree a -> a
addNums Empty = 0
addNums (Node n []) = n
addNums (Node n (x:xs)) = n + (addNums x) + addNums (Node 0 xs)

, n. , , bae (.. ).

ghci:

λ> addNums myNums
36

Answer 2

Foldable sum:

{-# LANGUAGE DeriveFoldable #-}

data Tree a = Empty | Node a [Tree a] deriving (Eq, Show, Foldable)

myNums :: (Num a) => Tree a
myNums = ...

main = print $ sum myNums

Answer 3

:

addNums :: (Num a) => Tree a -> a
addNums Empty = 0
addNums (Node n xs) = n + sum (map addNums xs)

xs. addNums , . n.