DS AVL Trees

Question

Consider the tree T in which left subtree contains half of the maximum number of nodes possible in the avl tree of height 6 and right subtree contains one 3rd of the minimum number of nodes possible in Avl tree of height 6.What will be total number of nodes in T?

 

Edit:- I am getting 75.5 as answer.Now i am not sure whether to pick 75 ot 76

Comments

yes $75.5$ coming.

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75

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How can u post a solution

Answer 1

Number of nodes in left subtree=(2^n-1)/2 =(2^6-1)/2=63

Number of elements in right subtree= (minimum number of nodes in avl at height 6)/3=33/3=11

Total number of nodes =1(root)+63+11=75

Answer 2

maximum number of nodes (left side) with height(h)=2^(h)-1 = 63

Now the minimum number of nodes N(h)=N(h-1)+N(h-2)+1 , N(0)=1, N(1)=2

 After solving the equation you will get N(6)=33 but we will see in only right most so

right side =33*1/3 = 1½ = 5

so total nodes= 63+5+1 =69