Final answer:
For the node at location 4999 in an array-based complete binary tree with 10,000 nodes, the right child would theoretically be at location 10000 using the formula (2i + 2). However, this exceeds the array bounds, indicating that the node at 4999 does not have a right child.
Step-by-step explanation:
Considering a complete binary tree with exactly 10,000 nodes implemented with an array, if a node has its value stored at location 4999 in the array, the location of its right child can be determined by a specific pattern. In array-based implementation of a binary tree, for any node at index i, the index of its left child is given by 2i + 1 and the index of its right child is given by 2i + 2. Therefore, to find the index of the right child of a node at index 4999, we apply the formula 2i + 2 where i is 4999.
The calculation would therefore be 2 * 4999 + 2 = 9998 + 2 = 10000. However, since we only have 10,000 nodes, and array indices start at 0, the highest index we can have is 9999. This means that if the node is at position 4999, there cannot be a right child at position 10000, as this is outside the scope of the array's indices which indicates that the last element at index 9999 is the rightmost child in the bottom level of the tree. Hence, this suggests that the node at 4999 is one of the last nodes at the penultimate level of the binary tree and therefore does not have a right child within the current tree configuration.