A full binary tree seems to be a binary tree in which every node is either a leaf or has 2 children. I have been trying to prove that its height is O(logn) unsuccessfully. Here is my work so far:
I am considering the worst case of a full binary tree in which each right node has a subtree, and each left node is a leaf.
In this case:
$N = 2x - 1$
$H = x - 1$
I am going nowhere trying to prove that $H = O(\log(N))$
Furthermore, we know that leaves l is bounded by $h+1 <l<2^h$.
Internal nodes is bounded by $h<i<2^{h-1}$.
All this proves is that number of nodes $n=i+e$ is $<= 2^{h+1} - 1$ i.e. $\log(n) <= h$. But this does not take me anywhere closer to prove that $H = O(\log(n))$