Conceptual question on conditional expectations for numerically solving Backward SDEs

I am starting to study the field of Backward Stochastic Differential Equations (BSDE) and have a conceptual question on numerical techniques to solve them. BSDE are of the form: $$Y_t=F((B_s)_{0\leq s\leq T})+\int_t^Tf(s,Y_s,Z_s)\text{d}s-\int_t^TZ_s\text{d}B_s$$ where $$Y_T=F((B_s)_{0\leq s\leq T})$$ is the terminal condition, $$B_t$$ a Brownian Motion, $$Z_t$$ a $$\mathcal{F}_t$$-adapted process which ensures the solution to the equation is adapted (where the filtration is normally the natural filtration of $$B_t$$) and $$f(\cdot)$$ is the so-called generator of $$Y_t$$. By adaptedness of $$Z$$ we have equivalently: $$Y_t=\mathbb{E}[Y_T|\mathcal{F}_t]+\int_t^T\mathbb{E}[f(s,Y_s,Z_s)|\mathcal{F}_t]\text{d}s$$ Thus numerical schemes to solve this kind of BSDE normally involve conditional expectations. Discretizing, we get something like: $$Y_{t_i}^\pi=\mathbb{E}[Y_{t_{i+1}}^\pi|\mathcal{F}_{t_i}^\pi]+f(t_i,Y_{t_i}^\pi,Z_{t_i}^\pi)\Delta_{i+1}$$ where the Brownian Motion $$B$$ have been discretized into a random walk $$B^\pi$$; $$Y^\pi$$ and $$Z^\pi$$ are discretized versions of $$Y$$ and $$Z$$; $$\pi$$ is some partition of $$[t,T]$$; $$\mathcal{F}^\pi_{t_i}$$ is the filtration generated by the random walk $$B^\pi$$; and $$\Delta_{i+1}=t_{i+1}-t_i$$.

Now, my question is on the calculation of the conditional expectation of the form $$\mathbb{E}[Y_{t_{i+1}}^\pi|\mathcal{F}_{t_i}^\pi]$$. In Ma, Protter, San Martín and Torres (2002), we have the following statement at the bottom of page 305:

We remark that the conditional expectation w.r.t. to the discrete filtration $$(\mathscr{F}^{(n)})$$ can be computed explicitly as follows. We assume $$\Gamma$$ is an $$\mathscr{F}_{t_{k+1}}^{(n)}$$-mesurable random variable and we take the $$2^k$$ atoms corresponding to the trajectories of the martingale [i.e. random walk] $$M^{(n)}$$ in $$\mathscr{F}_{t_k}^{(n)}$$. Each atom in $$\mathscr{F}_{t_k}^{(n)}$$ splits into two atoms of $$\mathscr{F}_{t_{k+1}}^{(n)}$$ [by definition of a random walk of course]. Then we have: $$\mathbb{E}[\Gamma|\mathscr{F}_{t_k}^{(n)}](\omega)=\frac{1}{2}(a+b)$$ where $$a$$, $$b$$ are the values of $$\Gamma$$ in the two atoms of $$\mathscr{F}_{t_{k+1}}^{(n)}$$ coming from the corresponding atom in $$\mathscr{F}_{t_k}^{(n)}$$ containing $$\omega$$.

I am not sure I understand this paragraph and thus how the conditional expectations are computed. The way I see, this is how we do it:

1. Simulate the random walk $$B^\pi$$ on $$t=t_1$$, $$t_2$$, $$\dots$$ , $$t_n=T$$ therefore obtaining a trajectory $$\omega$$;
2. Compute the terminal value $$Y_T^\pi=Y_{t_n}^\pi=F((B_{t_i}^\pi)_{1\leq i\leq n})$$ from the trajectory $$\omega$$;
3. $$\Gamma$$ above is interpreted as $$Y_{t_n}^\pi$$. We know the trajectory $$\omega$$ of $$B^\pi$$ therefore we know a) the value of $$B_{t_{n-1}}^\pi$$, b) the value of $$B_{t_n}^\pi$$, and c) $$Y_{t_n}^\pi \in \{a,b\}$$. From a) and b) we know whether the random walk went up or down between $$t_{n-1}$$ and $$t_n$$ thus if for example $$Y_{t_n}^\pi=a$$ we can easily compute $$b$$. Once $$b$$ has been derived we compute: $$\mathbb{E}[Y_{t_n}^\pi|\mathcal{F}_{t_{n-1}}^\pi](\omega)=\frac{1}{2}(a+b)$$

Is my understanding of the computation of the conditional expectation, in particular step 3, accurate?