Asymptotic analysis of $T(n) = T(n/5) + T(4n/5) + \Theta(n)$
Your recurrence can be solved using the Akra–Bazzi theorem, a vast generalization of the master theorem. The result is $\Theta(n\log n)$.
You can also see this using a recursion tree. The root is labeled $n$. An internal vertex $m$ has children labeled $m/5$ and $4m/5$. Vertices are leaves if their labels is at most some arbitrary positive constant $C$, say $C=1$. The running time is proportional to the sum of all labels.
We can partition the vertices in the tree to levels according to their distance from the root. The first $\log_5 n$ levels of so are &quot;complete&quot;, that is, their labels sum to $n$, and there are $\log_{5/4} n$ levels in total, so the sum of labels is between $n\log_5 n$ and $n\log_{5/4} n$ levels. In particular, the complexity is $\Theta(n\log n)$.