0 votes 0 votes Let $T$ be a rooted ternary tree where each internal node of $T$ has a maximum of $3$ children. If root is at depth $0$, then maximum total number of vertices $T$ can have with depth $3$ is ______. Algorithms tbb-algorithms-2 numerical-answers + – Bikram asked May 26, 2017 edited Aug 20, 2019 by Counsellor Bikram 284 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 3 votes 3 votes For the tree to have maximum total no of vertices, all nodes must have 3 children So total no of nodes for depth 3 = 3^0 + 3^1 + 3^2 + 3^3 = 40 shraddha priya answered May 31, 2017 selected May 31, 2017 by Bikram shraddha priya comment Share Follow See 1 comment See all 1 1 comment reply commenter commenter commented Dec 5, 2019 reply Follow Share Maximum number of nodes with depth 3 are 3^3. So answer should be 27 1 votes 1 votes Please log in or register to add a comment.