foldl function in SML/NJ: Right-hand-side of clause doesn't agree with function result type

Question

I have a function named dfs which produces a list of all visited nodes in a graph represented by a list of tuples such as [(node1,node2,weight),....] and I get this error :

v.sml:72.7-79.16 Error: right-hand-side of clause doesn't agree with function result type [circularity]
  expression:  ''Z -> ''Z list
  result type:  ''Z list -> ''Z list
  in declaration:
    find_visited = (fn arg => (fn <pat> => <exp>))

uncaught exception Error
  raised at: ../compiler/TopLevel/interact/evalloop.sml:66.19-66.27
             ../compiler/TopLevel/interact/evalloop.sml:44.55
             ../compiler/TopLevel/interact/evalloop.sml:292.17-292.20
- 

Code :

fun succesors n e =
  List.map (fn (_, v, _) => v) (List.filter (fn (u, _, _) => n = u) e)

fun dfs graph start =
  let
    fun find_visited visited node =
      if not (List.exists (fn x => x = node) visited) then
        let
          val s = (succesors node graph)
        in
          foldl (fn (v, n) => (find_visited v n)) (node::visited)  s
        end
      else 
        visited
  in
    find_visited [] start
  end

Answer 1

foldl has the type ('a * 'b -> 'b) -> 'b -> 'a list -> 'b.

That is, the function to fold is list_element * result -> result, and (fn (v, n) => ... should be (fn (n, v) => ...

(This is the opposite order to OCaml's fold_left.)