-- Level Order Traversal of a labelled binary tree
-- Objective: Print each "level" of the tree on a separate line
--
-- NOTICE: This demo relies on tricu base library functions
-- 
-- We model labelled binary trees as sublists where values act as labels. We
-- require explicit not?ation of empty nodes. Empty nodes can be represented 
-- with an empty list, `[]`, which is equivalent to a single node `t`. 
--
-- Example tree inputs:
-- [("1") [("2") [("4") t t] t] [("3") [("5") t t] [("6") t t]]]]
-- Graph:
--       1
--      / \
--     2   3
--    /   /  \
--   4   5    6
--

label = (\node : head node)

left  = (\node : if (emptyList? node) 
  [] 
  (if (emptyList? (tail node)) 
    [] 
    (head (tail node))))

right = (\node : if (emptyList? node) 
  [] 
  (if (emptyList? (tail node)) 
    [] 
    (if (emptyList? (tail (tail node))) 
      [] 
      (head (tail (tail node))))))

processLevel = y (\self queue : if (emptyList? queue) 
  [] 
  (pair (map label queue) (self (filter 
    (\node : not? (emptyList? node)) 
      (lconcat (map left queue) (map right queue))))))

levelOrderTraversal_ = (\a : processLevel (t a t))

toLineString = y (\self levels : if (emptyList? levels) 
  "" 
  (lconcat 
    (lconcat (map (\x : lconcat x " ") (head levels)) "") 
    (if (emptyList? (tail levels)) "" (lconcat (t (t 10 t) t) (self (tail levels))))))

levelOrderToString = (\s : toLineString (levelOrderTraversal_ s))

flatten = foldl (\acc x : lconcat acc x) ""

levelOrderTraversal = (\s : lconcat (t 10 t) (flatten (levelOrderToString s)))

exampleOne = levelOrderTraversal [("1") 
                                 [("2") [("4") t t] t] 
                                 [("3") [("5") t t] [("6") t t]]]

exampleTwo = levelOrderTraversal [("1") 
                                 [("2") [("4") [("8") t t] [("9") t t]] 
                                        [("6") [("10") t t] [("12") t t]]] 
                                 [("3") [("5") [("11") t t] t] [("7") t t]]]

exampleTwo