algorithm - tree graph - find how many pairs of vertices, for which the sum of edges on a path between them is C -


i've got weighted tree graph, weights positive. need algorithm solve following problem.

how many pairs of vertices there in graph, sum of weights of edges between them equals c?

i thought of solutions thats o(n^2)

for each vertex start dfs , stop when sum gets bigger c. since number of edges n-1, gives o(n^2) solution.

but can better?

for undirected graph, in terms of theoretic asymptotic complexity - no, cannot better, since number of such pairs o(n^2).

as example, take 'sun/flower' graph:

g=(v[union]{x},e) e = { (x,v) | v in v } w(e) = 1 (for edges)

it easy see graph indeed tree.

however, number of pairs have distance of 2 (n-1)(n-2) in omega(n^2), , algorithm finds of them omega(n^2) in case.


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