Let T be a binary tree with n nodes that is realized with an array list, S, and let p be the level numbering of the nodes in T. Give pseudo-code descriptions of each of the methods root, parent, left, right, hasleft, hasRight, islnternal, isExternal, and isRoot.

Assuming that the nodes in the array list S are stored in level order (i.e., the root is at index 0, its children at 1 and 2, their children at 3, 4, 5, and 6, and so on), the pseudo-code descriptions for each of the methods are as follows:

1. root: Return the node at index 0 in the array list S.

root(S):
    return S[0]

2. parent: Return the parent of the node at index i in the array list S. If i is the root, return None.

parent(S, i):
    if i == 0:
        return None
    else:
        return S[(i-1)//2]

3. left: Return the left child of the node at index i in the array list S. If the node has no left child, return None.

left(S, i):
    left_child = 2*i + 1
    if left_child < len(S):
        return S[left_child]
    else:
        return None

4. right: Return the right child of the node at index i in the array list S. If the node has no right child, return None.

right(S, i):
    right_child = 2*i + 2
    if right_child < len(S):
        return S[right_child]
    else:
        return None

5. hasLeft: Return True if the node at index i in the array list S has a left child, and False otherwise.

hasLeft(S, i):
    return (2*i + 1) < len(S)

6. hasRight: Return True if the node at index i in the array list S has a right child, and False otherwise.

hasRight(S, i):
    return (2*i + 2) < len(S)

7. isInternal: Return True if the node at index i in the array list S has at least one child (i.e., it is not a leaf), and False otherwise.

isInternal(S, i):
    return hasLeft(S, i) or hasRight(S, i)

8. isExternal: Return True if the node at index i in the array list S has no children (i.e., it is a leaf), and False otherwise.

isExternal(S, i):
    return not isInternal(S, i)

9. isRoot: Return True if the node at index i in the array list S is the root of the tree, and False otherwise.

isRoot(S, i):
    return i == 0