Let $(\mathbb G,\Gamma)$ be a graph of groups. A $G$-path from $u_0$ to $u$ is $$g_0e_1g_1\cdots e_{n}g_n,$$ where $e_1\cdots e_{n}$ is a walk in $\Gamma$ from $u_0$ to $u$ and each $g_i\in G_{s(e_i)}$. We denote the set of all $\mathbb G$-paths from $u_0$ to $u$ by $\pi[u_0,u]$. Let $\mathbb F(\mathbb G,\Gamma)$ be the group generated by the vertex groups and the elements $e$ of edge $\Gamma$, subject to the relations $$\{s_e(g)e=et_e(g)\mid g\in G_e,e\in E(\Gamma)\},\{e^{-1}=\bar{e}\mid e\in E(\Gamma)\}$$

Recall that the fundamental group of $(\mathbb G,\Gamma)$ on the base point $u_0$ is the set of all elements $g_0e_1g_1\cdots e_{n}g_n$ in $\mathbb F(\mathbb G,\Gamma)$, where $e_1\cdots e_{n}$ is a closed walk from $u_0$ to $u_0$. In other words, the fundamental group of $(\mathbb G,\Gamma)$ is $\pi[u_0,u_0]$.

My question is regarding the universal cover (Bass-Serre tree) of $(\mathbb G,\Gamma)$.

In the book "Trees'' by Serre page 51, the universal cover is defined in the following way:

The vertices are the left cosets of the vertex groups in $\pi[u_0,u_0]$, i.e.

$$\bigcup_{u\in V(\Gamma)} \pi[u_0,u_0]G_u$$

However in a paper by Bass "Covering theory for graphs of groups ", he defined in the following way:

The vertices are $$\bigcup_{u\in V(\Gamma)} \pi[u_0,u]G_u$$

My question is that

Are they $\pi[u_0,u_0]$-isomorphic?or they are completely different?

If yes, I really wonder to know the $\pi[u_0,u_0]$-isomorphism map.