BuildTree Analysis(Approximate)
T(1) = 1
T(n) = 2T(n/2) + 1
T(n) = 2(2T(n/4)+1) + 1
T(n) = 4T(n/4) + 2 + 1
T(n) = 4(2T(n/8)+1) + 2 + 1
T(n) = 8T(n/8) + 4 + 2 + 1
T(n) = 2kT(n/2k) +
let 2k = n, log n = k
T(n) = nT(1) +
T(n) = ?(n)
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