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The node of a binary tree is called a single child if it has a parent but does not have a sibling. The root is by definition not considered a single child.

Let $T$ be a binary tree of size $n$, and let $k$ be the number of vertices in $T$ that are single children. Is it true that if $\frac{k}{n}\leq \frac{1}{2}$ then the height of $T$ is $O(\log n)$?

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    $\begingroup$ What do you think? $\endgroup$ – Yuval Filmus Oct 11 '20 at 17:48
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You can construct a tree with no single children which has height $(n-1)/2$: Path tree.

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