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the average height of a leaf in a binary tree with 128 leaves is at least

the average height of a leaf in a binary tree with 128 leaves is at least

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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Based on Theorem 4.4.2 P.165, the average height of a leaf in a binary tree with 128 leaves is at least

4.4 Binary Trees with (at Least) n! Leaves 165 The contrapositive of this second statement is as follows: If a Binary Tree has >= n leaves, then it has a height >= l(n). And, in particular: If a Binary Tree has >= n! leaves, then it has a height >= lg(n!). Since the Branching Diagram of comparisons for any Sorting Algorithm applied to an array of length n must have >= n! leaves, the Worst Case for this algorithm (which corresponds to reaching a highest leaf) must do >= Ig(n!) comparisons (of array entries). // But what about the average case? Can the Branching Diagram give us a bound // for it? Theorem 4.4.2: The average height of a leaf in a Binary Tree with n leaves is >= lg(n). Proof. Let T be some (fixed but arbitrary) Binary Tree with n leaves. // We'll prove that the leaves in T have an average height >= lg(n) in four stages. Let SHL(T) denote the Sum of the Heights of the Leaves in T , and let AHL(T) denote the Average of the Heights of the Leaves in T. Then, AHL(T) = SHL(T)/n. // Stage 1. If some vertex x, which is not the root, has only one vertex w below it, and we remove x but join the vertex u above x directly to w, we obtain a new tree T1, with the same number of leaves but at least one leaf is now one step closer to the root. -> u 71
Transcribed Image Text:4.4 Binary Trees with (at Least) n! Leaves 165 The contrapositive of this second statement is as follows: If a Binary Tree has >= n leaves, then it has a height >= l(n). And, in particular: If a Binary Tree has >= n! leaves, then it has a height >= lg(n!). Since the Branching Diagram of comparisons for any Sorting Algorithm applied to an array of length n must have >= n! leaves, the Worst Case for this algorithm (which corresponds to reaching a highest leaf) must do >= Ig(n!) comparisons (of array entries). // But what about the average case? Can the Branching Diagram give us a bound // for it? Theorem 4.4.2: The average height of a leaf in a Binary Tree with n leaves is >= lg(n). Proof. Let T be some (fixed but arbitrary) Binary Tree with n leaves. // We'll prove that the leaves in T have an average height >= lg(n) in four stages. Let SHL(T) denote the Sum of the Heights of the Leaves in T , and let AHL(T) denote the Average of the Heights of the Leaves in T. Then, AHL(T) = SHL(T)/n. // Stage 1. If some vertex x, which is not the root, has only one vertex w below it, and we remove x but join the vertex u above x directly to w, we obtain a new tree T1, with the same number of leaves but at least one leaf is now one step closer to the root. -> u 71
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