【在 h*****n 的大作中提到】 : 2^n-n in total, : take 1~n as inorder traversal, : recursion i as root node, T(i) = T(i-1) + T(n-i) ?
m*6
26 楼
BB对网上reload一直查得不紧不慢啊,我的paypal debit前两天也被停了
h*e
27 楼
是的,又想起了趣味数学。
【在 i***e 的大作中提到】 : right. Catalan numbers!!
h*n
28 楼
I had a typo, the recursion should be multiplications T(n) = \sum_i=0^(n-1) T(i)*T(n-1-i) catalan numbers is correct, the recursions shows how "catalan numbers" are calculated.
【在 i***e 的大作中提到】 : that's wrong! please google "catalan numbers"