Number of Ways to Assign Edge Weights I

Compute the maximum depth (number of edges) from root 1 to any deepest node in the tree, then count assignments of weights {1,2} on that root-to-deepest-node path that make the total sum odd. Because parity depends only on the number of 1s, the answer is the number of length-m binary sequences with an odd number of 1s, i.e., 2^{m-1} modulo 1e9+7 (m = max depth).