Show that number of pendant vertices in a binary tree is (n+1)/2 ,where n is the number of vertices in the tree.

The number of vertices of odd degree in an undirected graph is even,(N-1) is even.
 Therefore, n is odd.
Let p be the number of pendant vertices.

2+p+3(n- p-1) = 2(n-1)


or, 3n-2n-3+4 = 2p


or, 2p = n+1


or, P = (n+1)/2


Hence, the number of pendant vertices is (n+1)/2.

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