00:01
As per the question, we have to find the recursion tree for recurrence relation t of n is equal to 4t of n by 2 plus n.
00:18
So, we start with t of n with n as the scalar part.
00:25
We have 4 tn by 2, t of n by 2, t of n by 2, t of n by 2 and t of n by 2.
00:36
Now this t of n by 2 will be further divided into 4 t of n by 2 square, t of n by 2 square, t of n by 2 2 square t of n by 2 square same we do for the other two also other three also t of n by 2 square t of n by 2 square t of n by 2 square t of n by 2 square same we do for the third one t , t , t and t.
01:33
And find the last one t , t , t and t.
01:51
With the scalar here will be 2n.
01:56
N by 2, n by 2, n by 2, n by 2, 4 times will be 2n...