If I understand correctly, you have a tree with edges labelled by e1,...,en and vertices labelled by v0,...,vn,...v(n+m-1), where the last m >= 1 of these label the same (namely the exceptional) vertex. Relations among Brauer characters are given by: vi = sum of the ej, where ej runs through all the edges adjacent to vi. You would like to find an algorithm to solve this overdetermined system of equations for the ej. Here is one: pick your favourite label at the exceptional vertex and discard the rest (or pick your favourite Z-linear combination whose coefficients add up to 1). Then we are reduced to the case when m = 1. If an edge ej has a leaf vi as vertex, it's easy: we have ej = vi. Next pick an edge ej such that all the adjacent edges at one of its two vertices vi have already been 'lifted'. Then we can use the equation for vi to solve for ej. Inductively work your way further and further away from the leaves. This will give you what you wanted.

Incidentally, you get all other solutions by using the relations vn = ... = v(n+m-1) and sum over even vertices = sum over odd vertices (for this pick an arbitrary vertex v; then say that another vertex is even/odd if it has even/odd distance from v; the relation holds because each edge has precisely one even and one odd vertex).

Unfortunately I don't know about Green's lifting.

If I understand correctly, you have a tree with edges labelled by e1,...,en and vertices labelled by v0,...,vn,...v(n+m-1), where the last m >= 1 of these label the same (namely the exceptional) vertex. Relations among Brauer characters are given by: vi = sum of the ej, where ej runs through all the edges adjacent to vi. You would like to find an algorithm to solve this overdetermined system of equations for the ej. Here is one: pick your favourite label at the exceptional vertex and discard the rest (or pick your favourite Z-linear combination whose coefficients add up to 1). Then we are reduced to the case when m = 1. If an edge ej has a leaf vi as vertex, it's easy: we have ej = vi. Next pick an edge ej such that all the adjacent edges at one of its two vertices vi have already been 'lifted'. Then we can use the equation for vi to solve for ej. Inductively work your way further and further away from the leaves. This will give you what you wanted.

Incidentally, you get all other solutions by using the relations vn = ... = v(n+m-1) and sum over even vertices = sum over odd vertices (for this pick an arbitrary vertex v; then say that another vertex is even/odd if it has even/odd distance from v; the relation holds because each edge has precisely one even and one odd vertex).

Unfortunately I don't know about Green's lifting.

Update: I claim that the Brauer tree alone does not determine Green's lift. In particular there is no algorithm for Green's lift that uses only the Brauer tree. The point is that in some cases there is no lift at all that respects the symmetries of the tree.

Example: suppose that $G = S_3$, $p = 2$. The principal block has the cyclic 2-Sylow as defect group. It contains only the trivial representation. The tree thus has only one edge, and there is no exceptional vertex (e.g. as its multiplicity is $p-1 = 1$). More precisely the two characteristic zero representations of this block are the trivial character T and the sign character S (both reducing to the trivial). So all the lifts of the trivial mod p representation are of the form $nT-(n-1)S$ for some integer $n$, but none of these is symmetric w.r.t. transposition of the vertices T, S.