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 $H = O(\log(n))$
Height of a full binary tree
graph theory;combinatorics;binary trees;trees
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 $H\notin O(\log N)$.