Suppose I am making a red-black search tree, and in my right subtree, I have a black node, then a red node, and it has two black children, the black children further black childrens. As such a lemma has been made that red-black trees with $n$ internal nodes have height at most $2\log(n+1)$, would this proof still hold for such a black tree?
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Have a close look at the definition. – Raphael Oct 28 '12 at 10:56
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In fact, a red-black tree can have all black nodes. In that case the tree is completely balanced and has the required bound.
Hendrik Jan
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