Exercises from Chapter 5
                  
5.3
5.6
5.9

                                   Solutions

5.3 Base case: n = 0.  Then the tree consists of only one leaf, so
E = 0 and I = 0, and the theorem holds.

Induction step: assume n > 0 and the theorem holds for full binary
trees with n-1 internal nodes.  Take a tree T with n internal nodes.
Since n > 0, T has internal nodes. Take an internal node v of T that has
maximum depth d among all internal nodes.  Because v is
internal and T is full, v has two children, each of depth d+1.
Because v has maximum depth among internal nodes, its children are
leaves.

Let U be the full tree that results from removing the two children
of v, J = U's internal path length, and F = U's external path length.
Since v is a leaf in U, U has n-1 internal nodes. By the induction
assumption, F = J + 2(n-1).

E = F + 2(d+1) - d = F + d + 2 because of the depths of v's two
children and because v is no longer a leaf, and
I = J + d because of v. Therefore
E = F + d + 2
  = J + 2(n-1) + d + 2 by our induction assumption
  = I - d + 2(n-1) + d + 2
  = I +2n.

5.6 The main idea is to use a queue to hold the subtrees that will be printed:

template <class Elem>
void level(BinNode<Elem>* subroot) {
  AQueue<BinNode<Elem>*> Q;
  Q.enqueue(subroot);
  while(!Q.isEmpty()) {
    BinNode<Elem>* temp;
    Q.dequeue(temp);
    if(temp != NULL) {
      Print(temp);
      Q.enqueue(temp->left());
      Q.enqueue(temp->right());
} } }

5.9 (a) Since all nodes have the same structure, the overhead fraction
is 12/16 = 3/4.

(b) 8/24 = 1/3.

(c) Letting i be the number of internal nodes and l be the number of
leaves, the overhead fraction is (12i + 4l)/(20i + 12l). For large
full trees, this is aproximately 16/32 = 1/2.

(d) Approximately 8/16 = 1/2.