Oh! sorry my idea for internal nodes=leaves-1 is only for binary tree like at level 4 total leaves 2^4=32 so internal nodes=32-1=31 also can be checked as 16+8+4+2+1=31
there must be a shortcut for ternary also... by the way brijesh thanks for correcting me. so total internal nodes will be 40 as stated by brijesh On Wed, Aug 10, 2011 at 3:08 AM, Brijesh Upadhyay < [email protected]> wrote: > General approach would be , get the no of levels first by log 28 /log 3 , = > 4(use ceiling)...and now 3^0+3^1+3^2+3^3 = 40 will be no of internal nodes.. > > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To view this discussion on the web visit > https://groups.google.com/d/msg/algogeeks/-/OyhD3tQ5uPgJ. > > To post to this group, send email to [email protected]. > To unsubscribe from this group, send email to > [email protected]. > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > -- **Regards SAGAR PAREEK COMPUTER SCIENCE AND ENGINEERING NIT ALLAHABAD -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
