Answer is 2N-4
binary tre having 2n-1 routers
In Nth level of tree each node(router) will take n-1 hops to tranfer to root and half of the total routers are on this level. (n-1)*1/2 hops
N-1 level 's nodes will take n-2 hops to tranfer to root and half of the remaining routers are on this level. (n-2)*1/4 hops
N-2 level 's nodes will take n-3 hops to tranfer to root and half of the remaining routers are on this level. (n-3)*1/8 hops
N-3 level 's nodes will take n-4 hops to tranfer to root and half of the remaining routers are on this level. (n-4)*1/16 hops
So in total i router to root
= (n-1)*1/2+ (n-2)*1/4+ (n-3)*1/8+ (n-4)*1/16+......
= n-2
Similarly root to router j again n-2
Mminimum no. Hops = 2(n-2)= 2n-4