# Height of a full binary tree

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 $H = O(log(n))$

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The height of the tree is bounded below logarithmically, but not above (in fact, you've shown that $H=\mathcal{O}(n)$). Are you sure the problem is right? This holds for a complete binary tree though. Is that what you meant? – EuYu Mar 13 '13 at 6:53
Your worst case shows explicitly that $h$ is not $O(\log n)$ for an arbitrary full binary tree. You need another assumption such as "all leaves have about the same depth" before it's true that $h = O(\log n)$. – Erick Wong Mar 13 '13 at 6:53
@Erick Good timing! :) – EuYu Mar 13 '13 at 6:55