Question

7. Find the overhead fraction for each of the Binary Tree implementations when: a) Both leaf nodes and internal nodes nodes store data and two child pointers. The data field requires eight bytes and each pointer requires four bytes. b) Assume you study a full Binary Tree with a very large number of nodes. Both leaf nodes and internal nodes nodes store data, and internal nodes store two child pointers. Pointers each require each four bytes. The data field requires 32 bytes.

Answer 1

Overhead Faction = total space taken by pointers/total space taken by data + total space taken by pointer Part a Conditions 1. Both the leaf nodes and the internal nodes store data and two child pointers. 2. Data requires 8 bytes and each pointer requires 4 bytes. Suppose, our tree has n number of nodes. Each node consumes 8 bytes for storing data and 2*4 i.e., 8 bytes for storing the pointer. So the Overhead Fraction is Overhead Fraction = 8*n/8*n + 8*n = 1/2 Part b Conditions: 1. The tree is Full Binary Tree with a large number of nodes (that means we can drop constants if any). 2. Both the leaf nodes and the internal nodes store data. Data requires 32 bytes. 3. Only the internal nodes store two child pointers. Pointers require 4 bytes each. We know, the number of leaf nodes in a Full Binary Tree is

Part a

Conditions

1. Both the leaf nodes and the internal nodes store data and two child pointers.

2. Data requires 8 bytes and each pointer requires 4 bytes.

Suppose, our tree has n number of nodes. Each node consumes 8 bytes for storing data and 2*4 i.e., 8 bytes for storing the pointer. So the Overhead Fraction is

Part b

Conditions:

1. The tree is Full Binary Tree with a large number of nodes (that means we can drop constants if any).

2. Both the leaf nodes and the internal nodes store data. Data requires 32 bytes.

3. Only the internal nodes store two child pointers. Pointers require 4 bytes each.

We know, the number of leaf nodes in a Full Binary Tree is one more than the total number of the internal nodes. If we have N number of leaf nodes, then we have N-1 internal nodes. Now, calculating the memory usages:

Total Number of Nodes = N + (N -1) = 2N -1

1. Data Memory Usage = (2N - 1)* 32

2. Pointer Mem. Usage = (N-1) * 8

As N is a very large number, we can approximate N-1 ~ N

Replacing it in the above equation, we get

Note - If this answer is helpful, please consider giving a thumbs up, your feedback will be truly appreciated.