$t(n) = t(\alpha n) + t(\beta n) + cn$ , $\alpha + \beta \lt 1, c > 0$ Using tree recursion

Question

Solve $$t(n) = t(\alpha n) + t(\beta n) + cn$$ $$ \alpha + \beta \lt 1, c > 0$$

Using tree recursion

I've taken this problem when $\alpha + \beta = 1, t(n)= t(n/3) + t(2n/3) + n$

And I've got that it is $\Theta(nlogn)$

Can anyone explain how can I solve this when $\alpha + \beta < 1$ ? Maybe there's some reference in the internet already for this..

@AlexR. I'm studying right now the Master Theorem. It's kinda wierd and I'm not sure how to apply it because it says here $T(n) = a \cdot T(\frac{n}{b}) + f(n)$. but I have twice $t$ in my right side of the equation, how can I use $a$ and $b$ properly to prove it? — Ilan Aizelman WS, Aug 15 '16 at 19:59
The result is $\Theta(n)$. I proved this here: https://math.stackexchange.com/questions/506489/if-tn-un-sum-i-t-lfloor-r-i-n-rfloor-show-that-tn-thetan — marty cohen, Aug 15 '16 at 20:00
@martycohen I just found here, in the bottom of the page that it says $\Theta(n)$. https://www.ics.uci.edu/~eppstein/161/960130.html , and I find hard to understand your proof. I'll stick to Master's theorem I guess. I'm trying to figure how to to use it though. Will update later — Ilan Aizelman WS, Aug 15 '16 at 20:00

score 0 · Accepted Answer · answered Aug 17 '16 at 12:09

Building a tree:

We have to compute the work done:

$n$ - first line, root

$n/a + n / b$ - second line, children of root (sum of it $n/a + n/b$)

and like that so on..

Then we sum all the lines(because we need to sum all the work done), and we'll see that we will never exceed $n$.

So the answer is $\Theta(n)$

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