let n, e be the total no. of nodes and edges of the binary tree respectively. let n0 = total no. of nodes with 0 children n1 = total n0. of nodes with 1 child n2 = total no. of nodes with 2 children therefore, n = n0 + n1 + n2......(1) again e= n-1.....................(2) also e = n1 + 2*n2...............(3) from (2) and (3) n-1=n1 + 2*n2 =>n=1 + n1 + 2*n2.............(4) from (1) and (4) n0 + n1 + n2 = 1 + n1 + 2*n2 =>n0=1 + n2... proved