How does a disease spread through a triangular network? Consider a population of nodes arranged in a triangular configuration as shown in the figure below, where each level $k$ has $k$ nodes. Each node, except the ones in the last level, is a parent node to two child nodes. Each node in levels $2$ and below has $1$ parent node if it is at the edge, and $2$ parent nodes otherwise.
The single node in level $1$ is infected (red). With some probability $p_0$, it does not infect either of its child nodes in level $2$. With some probability $p_1$, it infects exactly one of its child nodes, with equal probability. With the remaining probability $p_2=1-p_0-p_1$, it infects both of its child nodes.
Each infected node in level $2$ then acts in a similar manner on its two child nodes in level $3$, and so on down the levels. It makes no difference whether a node is inected by one or two parents nodes - it's still just infected.
The figure below shows one possibility of how the disease may spread up to level $6$.

The question is: what is the expected number of infected nodes at level $k$?
Simulations suggest that this is (at least asymptotically) linear in $k$, i.e.,
$$
\mathbb{E}(\text{number of infected nodes in level } k) = \alpha k
$$
where $\alpha = f(p_0, p_1,p_2)$.

This question arises out of a practical scenario in some research I'm doing. Unfortunately, the mathematics involved is beyond my current knowledge, so I'm kindly asking for your help. Pointers to relevant references are also appreciated. 
I asked a different version of this question some time ago, which did not have the possibility of a node not infecting either of its child nodes. It now turns out that in the system I'm looking at, the probability of this happening is not negligble.
 A: We can easily give an asymptotic solution to this problem. Assume the probability that a node in the row $k$ is infected converges to a stable limit $\alpha$ (i.e. it will be the same for each $k\gg1$). Let $x$ be a node in the row $k+1$. Let
$$\beta := \frac{1}{2}p_1+p_2,$$
the probability that an infected parent of $x$ infects $x$. If exactly one parent of $x$ is infected, the probability that $x$ is infected is $\beta$, and if both parents are infected, it is $1-(1-\beta)^2 = 2\beta - \beta^2$.
Therefore:
\begin{align}
\alpha =&\ P(x\text{ is infected})\\
=&\ P(\text{exactly one parent of $x$ is infected})\beta\\&+ P(\text{both parents of $x$ are infected})(2\beta-\beta^2)\\
=&\ 2\alpha(1-\alpha)\beta + \alpha^2(2\beta-\beta^2).
\end{align}
Reordering terms, we obtain the equation
$$-\alpha^2\beta^2 + \alpha(2\beta - 1) = 0.$$
This has the solutions $\alpha_1=0$ and
$$\alpha_2=\frac{2\beta-1}{\beta^2}.$$
Notice that $\alpha_2\in[0,1]$ if, and only if $\beta\in[\tfrac{1}{2},1]$. This would correspond philosophically to the fact that $\alpha_2$ is a "stable" solution only for $\beta\in[\tfrac{1}{2},1]$, and else $\alpha_1$ is stable.
Can you check to see if you get the same result numerically?
Remark: As remarked in the comments, the probability that a a node in a row is infected depends on the position of the node in the row. The solution I presented ideally approximates the behavior in the middle of the row, where the situation is similar to starting with an infinite row instead of a single node and extending from there.
A: Interesting problem. I will address the case where each infected node independently infects its two children, each with probability $q$. As discussed in the comments, this is not the same as your stated problem, but since you expressed interest in this problem too, I'll attempt a solution. 
My general idea is that if we can find the probability for each node to be infected then we can calculate all other interesting quantities, including the expected value of infected nodes at level $k$.
I started by noticing that a node can be infected via many equal length "paths". The number of paths that exist from the root to any node is a binomial coefficient, and if we were to write the number of all possible paths from root to each node we would end up with Pascal's triangle. I noticed that each of these paths has the same probability to happen, equal to $q^n$, where $n$ is the length of the path. My first thought was that the probability of a node being infected was the number of paths times the probability of a single path. But this is not correct, since the paths are not independent to each other. There is a lot of overlap of the paths, so they are highly dependent. Then I decided to calculate the infection probability recursively, based on the probability of the parents being infected. A node is infected if either parent is infected and it also succesfully transmits the disease to the node of interest, which happens with probability $q$. And conversely, a not is not infected, if none of the parents are infected, or if one or two parents are infected but they do not transmit the disease. 
Let $P^k_i)$ be the probability of the $i^{th}$ node at level $k$ being infected.
Note that the left parent of this node is node $i-1$ at level $k-1$ and the right parent of this node is node $i$ at level $k-1$. Each level $k$, starts with node $1$ and ends with node $k$. If a node does not exist, such as node $0$, or node $k+1$ at level $k$ then we say that it's probability of infection is $0$.
This is how $P^k_i$ is defined recurcively:
$$
\begin{array}{rlr}
P^k_i = 1- (&(1-P^{k-1}_{i-1})\cdot(1-P^{k-1}_{i}) \quad &+ \\
& (1-P^{k-1}_{i-1})\cdot P^{k-1}_{i} \cdot (1-q) \quad &+ \\
& P^{k-1}_{i-1}\cdot (1-q) \cdot(1-P^{k-1}_{i}) \quad &+ \\
& P^{k-1}_{i-1}\cdot (1-q) \cdot P^{k-1}_{i}\cdot (1-q) )
\end{array}
$$
We can recursively calculate all $P^k_i$ for a large number of levels quickly. Let $N_k$ be the number of infected nodes at level k. The expected number of infected nodes at a level $k$ is simply
$$\mathbb E(N_k) = \sum_{i=1}^k P^k_i$$
I wrote a small python program to calculate this for different q. The results are quite interesting. The relationship is not linear with $k$. It looks linear for most values of $q$, but this is just deceptive. Here's a graph of the expected number of infected nodes for $q$ that show the non-linear relationship. 

I would like to find a closed form formula for $P^k_i$, but I need to study some background material for this. My current understanding is that since the recursive relationship is not linear, we might not be able to find a closed form.
