TAOCP 2.3.4.2 Exercise 14
The root of $G'$ is $V_1$, since the arcs of the oriented tree are $e_{01}$ and $e_{21}$.
Section 2.3.4.2: Oriented Trees
Exercise 14. [**] $$21$$ Let $G$ be the balanced digraph shown in Fig. 36, and let $G'$ be the oriented subtree with vertices $V_0,V_1,V_2$ and arcs $e_{01},e_{21}$. Find all oriented walks $P$ that meet the conditions of Theorem D, starting with arc $e_{12}$.
Verified: yes
Solve time: 1m07s
The root of $G'$ is $V_1$, since the arcs of the oriented tree are $e_{01}$ and $e_{21}$. By Theorem D, $e_{01}$ must be the last exit from $V_0$, and $e_{21}$ must be the last exit from $V_2$. Since the walk is required to start with $e_{12}$, the next exit from $V_2$ cannot be $e_{21}$; therefore it must be $e_{20}$. After reaching $V_0$, the next exit cannot be $e_{01}$; therefore it must be $e_{02}$. At this point all other exits from $V_2$ have been used, hence the next arc is $e_{21}$. Arriving at $V_1$, the only unused exit is $e_{10}$, and after reaching $V_0$ the only unused exit is $e_{01}$.
Thus the walk is forced to be
$$ P=(e_{12},e_{20},e_{02},e_{21},e_{10},e_{01}). $$
This walk uses every arc exactly once, begins with $e_{12}$, and has last exits $e_{01}$ from $V_0$ and $e_{21}$ from $V_2$. Hence it satisfies the conditions of Theorem D. No alternative choice is possible at any stage, so this is the only such walk.
$$ \boxed{P=(e_{12},e_{20},e_{02},e_{21},e_{10},e_{01})}. $$