The zero-sum tree-factor conjecture

Let $k$ and $n$ be positive integers such that

$${n\choose 2}\quad\text{and}\quad\frac{(k-1)n}{k}$$

are both even integers. Let $T$ be a tree of order $k$, and let $c:E(K_n)\to\{-1,1\}$ be a zero-sum labeling of the complete graph $K_n$, meaning that

$$c(E(K_n))=\sum_{e\in E(K_n)}c(e)=0.$$

Assume that $n$ is sufficiently large in terms of $k$. The zero-sum tree-factor conjecture. The graph $K_n$ has a zero-sum $T$-factor: there is a spanning forest $F$ of $K_n$ whose components are all isomorphic to $T$ and satisfy

$$c(E(F))=0.$$

The divisibility conditions are necessary for a zero-sum $T$-factor, and the conjecture seeks their sufficiency for every fixed tree when $n$ is sufficiently large relative to its order. The source presents this as an open conjecture motivated by the corresponding perfect-matching question.