I have this tree[1]:
(tree
'(game A B)
(tree '(game A C) '() (tree '(game C B) '() (tree '(game B A) '() '())))
(tree '(game B C) '() (tree '(game C A) '() (tree '(game A B) '() '()))))
This is how I'd like to print it out:
;; (game A B) --> (game A C) --> end
;; (game A B) --> (game A C) --> (game C B) --> end
;; (game A B) --> (game A C) --> (game C B) --> (game B A) --> end
;; (game A B) --> (game B C) --> end
;; (game A B) --> (game B C) --> (game C A) --> end
;; (game A B) --> (game B C) --> (game C A) --> (game A B) --> end
You see, some branches have no left or right nodes. In such cases,
I'd like to avoid going down the tree, otherwise I end up displaying
two different paths as the same path.
To do that, I had to put a second base case in the function below.
(So I'm calling displayln in two different places.) It looks ugly and
I wonder how you would do it. Would you do it differently?
(define (print-tree acc-str tree)
(cond [(empty? tree) (displayln (string-append acc-str " end"))]
;; avoid going down the tree if both branches are empty
[(and (empty? (tree-left tree))
(empty? (tree-right tree)))
(displayln (string-append acc-str (format "~a " (tree-root
tree)) "--> end"))]
[else
(print-tree
(string-append acc-str (format "~a " (tree-root tree)) "--> ")
(tree-left tree))
(print-tree
(string-append acc-str (format "~a " (tree-root tree)) "--> ")
(tree-right tree))]))
Thank you!
(*) Footnotes
[1] This tree represents the following tourney. A matches B. If A
wins, it matches C. If B wins, B matches C. In general, a winning
player matches the third player (who is not playing). If a player
wins two games in a row, the tourney is over. Also, if 4 matches in a
row take place, the tourney is over.
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