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A binary tree is \emph{full} if every non-leaf node has exactly two children. For context, recall that we saw in lecture that a binary tree of height $h$ can have at most $2^{h+1}-1$ nodes and at most $2^h$ leaves, and that it achieves these maxima when it is \emph{complete}, meaning that it is full and all leaves are at the same distance from the root. Find $\nu(h)$, the \emph{minimum} number of leaves that a full binary tree of height $h$ can have, and prove your answer using ordinary induction on $h$. Note that tree of height of 0 is a single (leaf) node.       \textit{Hint 1: try a few simple cases ($h = 0, 1, 2, 3, \dots$) and see if you can guess what $\nu(h)$ is.}

A binary tree is \emph{full} if every non-leaf node has exactly two children. For context, recall that we saw in lecture that a binary tree of height $h$ can have at most $2^{h+1}-1$ nodes and at most $2^h$ leaves, and that it achieves these maxima when it is \emph{complete}, meaning that it is full and all leaves are at the same distance from the root. Find $\nu(h)$, the \emph{minimum} number of leaves that a full binary tree of height $h$ can have, and prove your answer using ordinary induction on $h$. Note that tree of height of 0 is a single (leaf) node.       \textit{Hint 1: try a few simple cases ($h = 0, 1, 2, 3, \dots$) and see if you can guess what $\nu(h)$ is.}

Database System Concepts
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
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A binary tree is \emph{full} if every non-leaf node has exactly two children. For context, recall that we saw in lecture that a binary tree of height $h$ can have at most $2^{h+1}-1$ nodes and at most $2^h$ leaves, and that it achieves these maxima when it is \emph{complete}, meaning that it is full and all leaves are at the same distance from the root. Find $\nu(h)$, the \emph{minimum} number of leaves that a full binary tree of height $h$ can have, and prove your answer using ordinary induction on $h$. Note that tree of height of 0 is a single (leaf) node. 
    
\textit{Hint 1: try a few simple cases ($h = 0, 1, 2, 3, \dots$) and see if you can guess what $\nu(h)$ is.}

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