Lemma 14.1

A RB tree with n internal nodes has height at most 2*lg(n+1).

Thus the dynamic set operations on RB trees are all O(lgn).

Proof:

1.

First show that subtree rooted at x contains at least 2^bh(x) - 1 internal nodes.

Proof by induction.

Initial condition: if height(x) = 0, then x is a leaf whose subtree contains at least 2^bh(x) - 1 = 2⁰ - 1 = 0 internal nodes.

Inductive Step: Consider internal node x. Each child has black-height bh(x) (if the child is Red) or bh(x)-1 (if the child is Black).

By the Inductive Hypothesis, the child has at least 2^bh(x)-1-1 internal nodes.

Therefore the subtree rooted at x has at least (2^{bh(x) - 1} - 1) + (2^{bh(x) - 1} - 1) + 1 internal nodes, or 2^bh(x) - 1 internal nodes.

2.

Next, by property 3, a tree of height h has a black-height of at least h/2.

bh(T) $\geq$ h/2

n $\geq$ 2^bh(T) - 1

n $\geq$ 2^h/2 - 1, n+1 $\geq$ 2^h/2

lg(n+1) $\geq$ lg(2^h/2)

lg(n+1) $\geq$ h/2

h $\leq$ 2lg(n+1)