\[le:unidirectional-morph\] Let $\Gamma_0$ and $\Gamma_1$ be two order-preserving strictly-upward straight-line planar drawings of a rooted ordered tree $T$. Suppose that, for each node $v\in V(T)$, we have $y_{\Gamma_0}(v)=y_{\Gamma_1}(v)$. Then the linear morph $\langle \Gamma_0,\Gamma_1 \rangle$ is planar.

The proof exploits Corollary 7.2 in [@DBLP:journals/siamcomp/AlamdariABCLBFH17], which we introduce in the following.

Assume that the morph $\langle \Gamma_0,\Gamma_1 \rangle$ happens between the time instants $t=0$ and $t=1$. For any $t\in [0,1]$, denote by $\Gamma_t$ the drawing of $T$ in $\langle \Gamma_0,\Gamma_1 \rangle$ at time $t$ and denote by $u_t$ the position of a node $u$ at time $t$, where $u_t=(1-t) \cdot u_0 + t \cdot u_1$.

Now, consider a point $q_0$ *of* $\Gamma_0$, that is a point that represents a node of $T$ or that belongs to a segment representing an edge of $T$ in $\Gamma_0$. The morph $\langle \Gamma_0,\Gamma_1 \rangle$ moves $q_0$ to a point $q_1$ of $\Gamma_1$. This is evident if $q_0$ represents a node of $T$ in $\Gamma_0$; if $q_0$ is a point of the segment representing an edge $(u,v)$ in $\Gamma_0$, then $q_0=(1-\gamma) \cdot u_0+ \gamma \cdot v_0$, for some value $0<\gamma<1$, and the point $q_1=(1-\gamma) \cdot u_1 + \gamma \cdot v_1$ indeed belongs to the segment representing 