"no of possible minheaps" Code Answer's
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no of possible minheaps
VeNOM
on Oct 18, 2020
T(N) =(N-1)Ck * T(k) * T(N-k-1), where k = number of nodes on left subtree
T(1) = 1
T(2) = 1
T(3) = 2
T(4) = 3C2 * T(2) * T(1) = 3
T(5) = 4C3 * T(3) * T(1) = 8
T(6) = 5C3 * T(3) * T(2) = 20
T(7) = 5C3 * T(3) * T(3) = 80
Source: gateoverflow.in
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