GATE | GATE-CS-2015 (Set 2) | Question 20
A binary tree T has 20 leaves. The number of nodes in T having two children is
(D) Any number between 10 and 20
Sum of all degrees = 2 * |E|.
Here considering tree as a k-ary tree :
Sum of degrees of leaves + Sum of degrees for Internal Node except root + Root's degree = 2 * (No. of nodes - 1). Putting values of above terms, L + (I-1)*(k+1) + k = 2 * (L + I - 1) L + k*I - k + I -1 + k = 2*L + 2I - 2 L + K*I + I - 1 = 2*L + 2*I - 2 K*I + 1 - I = L (K-1)*I + 1 = L Given k = 2, L=20 ==> (2-1)*I + 1 = 20 ==> I = 19 ==> T has 19 internal nodes which are having two children.
This solution is contributed by Anil Saikrishna Devarasetty
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