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 $H = O(\log(n))$

  • 6
    $\begingroup$ Perhaps you mean a complete binary tree? The names are alike, but a complete tree is balanced, and certainly has the $\log$ property. $\endgroup$ Commented Mar 13, 2013 at 8:24
  • $\begingroup$ Yes, there are many different names around that are sometimes used interchangeably, especially when texts are translated. $\endgroup$
    – Raphael
    Commented Mar 13, 2013 at 14:40
  • $\begingroup$ I explained it here: stackoverflow.com/a/13093274/550393 $\endgroup$ Commented Apr 10, 2014 at 15:41

1 Answer 1


Your claim is incorrect (which might make it really hard to prove...) Indeed, as you describe, you can have a full binary tree of height $O(n)$: Let every right child be a leaf, and every left child have 2 children, until some level in which it has two leaf-children.

It holds that $x-1\in \theta(2x-1)$, and in particular, $x-1\in \omega(\log(2x-1))$, so $H\notin O(\log N)$.


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