= Solution
Follow the <binary decision diagram> from its top <decision node> to a <terminal node>, choosing the outgoing arrow specified by each tested <Boolean variable>. If $x=0$, the left $y$ <decision node> returns $y$, independently of $z$. If $x=1$, the right $y$ <decision node> returns one when $y=1$ and otherwise returns $z$. Thus the <truth table>, with rows in the order $(x,y,z)$, is:
|| $x$
|| $y$
|| $z$
|| $f(x,y,z)$
| 0
| 0
| 0
| 0
| 0
| 0
| 1
| 0
| 0
| 1
| 0
| 1
| 0
| 1
| 1
| 1
| 1
| 0
| 0
| 0
| 1
| 0
| 1
| 1
| 1
| 1
| 0
| 1
| 1
| 1
| 1
| 1
Thus the output column of the <truth table>, in the displayed order, is $\boxed{(0,0,1,1,0,1,1,1)}$. In particular, the skipped $z$ test on the left does not constrain $z$: both values give the same result.
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