On Sun, Aug 5, 2018 at 3:40 PM Matthias Felleisen
<matth...@felleisen.org> wrote:
> I“d write it like this:
>
> (struct tree (val left right) #:transparent)
> ;; Tree = '() | (tree X Tree Tree)
>
> (define (printer t0)
> (local (;; from-t0 : the path from t0 to t
> (define (printer/acc t from-t0)
> (cond
> [(empty? t)
> (map display (reverse (cons " end" from-t0)))
> (newline)]
> [else
> (define from-t (list* " --> " (tree-val t) from-t0))
> (printer/acc (tree-left t) from-t)
> (unless (empty? (tree-right t))
> (printer/acc (tree-right t) from-t))])))
> (printer/acc t0 '())))
>
> (printer tree1)
I took a while to understand this. It's brilliant. (Thanks for
taking the time --- no matter how little that might've been for you.)
FWIW, to understand it, I had to use the substitution method on paper.
Here's my final version --- using string-join. Thanks again!
(require racket/local)
(require racket/string)
(define (print-tree-elegant t0)
(local ((define (format-all-items ls)
(map (lambda (item) (format "~a" item)) ls))
(define (printer/acc t path-from-t0-to-t)
(cond
[(empty? t)
(displayln
(string-join
(reverse (format-all-items path-from-t0-to-t)) " --> "))]
[else
(define from-t (cons (tree-root t) path-from-t0-to-t))
(printer/acc (tree-left t) from-t)
(unless (empty? (tree-right t))
(printer/acc (tree-right t) from-t))])))
(printer/acc t0 '())))
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