Question

Assume that for some binary tree node implementation, a pointer requires 8 bytes and a data object requires 12 bytes. Further assume that we are storing a Full binary tree, that the internal nodes are implemented to store two pointers and a data field, and that the leaf nodes store only a data field. Type a fraction (like "1/2") that represents the fraction of the total space taken up by overhead. Give your fraction in lowest terms.

Answer 1

The fraction of the total space taken up by overhead in a full binary tree can be computed by dividing the total overhead by the total memory used by internal and leaf nodes. This involves accounting for both the pointers in internal nodes and the data fields in all nodes.

Step-by-step explanation:

To calculate the fraction of the total space taken up by overhead in a full binary tree, we need to consider the memory usage of pointers and data fields in internal and leaf nodes. Internal nodes in a full binary tree contain two pointers (each 8 bytes) and one data field (12 bytes), leading to a total of 28 bytes per internal node. Leaf nodes, however, consist only of a data field, which is 12 bytes. In a full binary tree, the number of leaf nodes is one more than the number of internal nodes.

Let's denote the number of internal nodes as n. Then, there would be n + 1 leaf nodes. The total memory for internal nodes is n * 28 bytes, and for leaf nodes, (n + 1) * 12 bytes. The overhead consists of the pointers, which is n * 2 pointers * 8 bytes/pointer. To find the fraction taken by overhead, we can express it as:

Total Overhead / Total Memory = (8 bytes/pointer * 2 pointers/node * n internal nodes) / (n internal nodes * 28 bytes/internal node + (n + 1) leaf nodes * 12 bytes/leaf node).

By simplifying the fraction, we can remove the common factor of n and reduce it to its lowest terms to get the exact fraction that represents the space taken by overhead in the full binary tree.