Solving the complexity of recursive time

Method a: Method substitution (substitution method)

This method is divided into two steps:
1,I guess the answer, We only need to guess substantially form
such as T (n) = 4T (n / 2) + n
by observing the recursion formula, n is doubled when noted, the output of four-fold increase, so speculation that the time complexity is recursion O (n- ² ), i.e., T (n-) = O (n- ² ).
2,Mathematical induction to prove
We constant coefficient of expansion tape
T (K) ≤ C _{. 1} K ² -C ₂ K (K <n-)
T (n-) ≤4 [C _{. 1} (n-/ 2) ² -C ₂ (n-/ 2)] + C = n- _{. 1} n- ² + (1-2c ₂ ) n-C = _{. 1} n- ² -C ₂ N- (C ₂ -1) n≤c _{. 1} n-2-C ^ ₂ n-
here we get a time complexity of O (n- ² )

So why not CK ² it, fancy style can not see, you can deduce what, simplified to CN ² + the n-, this formula we get T (the n-) ≤cn ² , it is achieved by reduced order ( calculate the time complexity we consider only the higher-order, do not control the low-order)

Method two: a recursive tree method (iterative method)

This formula should better understand it, is to unfold one by one, several geometric sequence mode to growth (d ^k-1 , each recursively d times, k is recursive layers), then here you can understood as a full d-tree .