Suppose $\Gamma$ is a (necessarily connected) weighted tree with $k$ vertices $v_1,\dots,v_k$. Suppose the weight of $v_i$ is $n_i\in \Bbb Z$. Let $A$ be the $k\times k$ symmetric matrix defined by $a_{ii}=n_i$ and for $i\neq j$, $$a_{ij}=\begin{cases}1 & \textrm{if there is an edge from $v_i$ to $v_j$}, \\ 0 & \textrm{otherwise} \end{cases}$$ The matrix $A$ defines a lattice, i.e. a symmetric bilinear form over $\Bbb Z$, on $\mathbb{Z}^k$, by $(x,y)\mapsto x^tAy$. A lattice $(\Bbb Z^n, Q)$ is said to be decomposable if $(\Bbb Z^m,Q)\cong (\Bbb Z^{m_1},Q_1)\oplus (\Bbb Z^{m_2},Q_2)$ for some lattices $(\Bbb Z^{m_i},Q_i)$ with $m_i\geq 1$, and indecomposable otherwise.
My question is: Is the lattice $(\mathbb{Z}^k,A)$ defined above indecomposable? I can't see whether this is true or not, but I'm guessing that if it is true, then the connectivity of $\Gamma$ should be used (because if $\Gamma$ is not connected then the corresponding lattice is clearly decomposable).
P.S. Maybe the assumption that $\Gamma$ is a tree can be replaced by the assumption that $\Gamma$ is merely a connected graph.