This feels like a homework problem, but the minimum weight would be:
Given the case:
n=1, the binary tree would be S the singleton set, so just a leaf with a weight of 1. When n=1 the weight is always 1, so minimum weight is n.
The maximum weight would be:
Given the fact that:
Binary trees are recursive, shown in the format of Left, Singleton, Right, so we can break any binary tree down to an n-size of 1 (L and R are empty sets, plus the Singleton set), 2 (Either L or R is an empty set, but not both), or 3 (Neither L nor R are empty sets). From this, when n is 1 we've shown that max weight is n (Which is also 2n-1 for this specific case). When n is 2 it must be a node with a single child, so weight is 2+1=3 (2n-1). When n is 3 it's a node with two children, each are a leaf, so 3+1+1 (also 2n-1).
Due to the recursive nature of Binary Trees, this can be propagated up to any value of n.