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Prove properties by induction on k

Base Case: Holds for B0.



Assume: Holds for \(B_0 \ldots B_{k-1}\).

1.
Bk is 2 copies of Bk-1, so \(2^{k-1} \;+\; 2^{k-1} \;=\;
2^k\) nodes.
2.
Depth of Bk is one greater than maximum depth of Bk-1.
Add one more level: height = (k-1) + 1 = k.
3.
See book (Lemma 19.1).
4.
True for children Bk-1, Bk-2, .., B0 from left to right.
Bk-1 is left child of Bk, root is also root of Bk-1 (minus left child), so degrees are Bk-1, Bk-2, .., B0.
The root of Bk is a Bk-1 with one more child (the left child), so root of Bk has degree (k-1) + 1 = k.


next up previous
Next: Up: Previous: BK