Option (C)
let T be the total nodes
$T=n+i $ n=leaves,i=internal nodes;
so
$i = T-n$
Degree=degree of leavers + degree of internal nodes
$D=1*n+3*(i-1)+2$
i-1 because all other internal nodes except root has degree 3 ,root has degree 2
$D=2e$ e= edges
also $e=T-1$
so
$2(T-1) = n+3i-3+2$
$2T-2 = n+3i-1$
putiing value odf i $i=T-n$
$2T-2 = n+3(T-n)-1$
$2T-2 = n+3T-3n-1$
$2T = 3T-2n-1+2$
$2T = 3T-2n-1+2$
$2T = 3T-2n+1$
$T = 2n -1$
Option (C)