Here we can use sum of degree= 2* number of edges
Now in a terneray tree we have root of degree 3 , non root internal node of degree 4 ( 1 indegree+3 outdegree) and leaves of degree 1(indegree as 1).
so sum of degree = 3R + 4Z +1L where R=root , Z=non root internal nodes, L =leaves . Also R=1 here
sum of degree = 3+4Z+L --------------------------------(1)
Now in a tree ,no of edges= number of nodes -1=[1 ( for root) + Z + L ]-1 =Z+L
|E| = Z + L ------------------------------(2)
substituting 1 and 2 in above formula yields
3+4Z+L=2(Z+L)
==> 3+4Z+L-2Z=2L
==> L=2Z+3
I know this differ from the options . I cant find where i m going wrong in this question..Please review my solution .