Right, but evaluating tree-size is O(n), even with a binary search tree. Doing
this on each insertion is slow.
The following is a possible solution without "do", and it only requires O(log
n) on a (balanced) search tree:
(define tree5
(let loop ([tree null] [n 5])
(define x (random 10))
(cond
[(zero? n) tree]
[(contains? tree x) (loop tree n)]
[else (loop (insert tree x) (sub1 n))])))
But I thought there must be a more elegant solution using iterations and
comprehensions. Thank you anyway!
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