Terminology

Learn all these terms:

• subtree
• edge
• parent
• path
• path length
• ancestor
• descendent
• depth
• height
• level
• leaf
• internal
• full
• complete
• perfect

The Wonder and Joy of Binary Trees

• We will spend a lot of time thinking about binary trees.
• Why????
• Question for you:

  • What is the maximum number of nodes in a binary tree of height $h$?

  • $N(h)\le ??$

    ${\displaystyle \sum _{i=1}^{n}c=cn}$	${\displaystyle \sum _{i=1}^{n}i={\displaystyle \frac{n(n+1)}{2}}}$
    ${\displaystyle \sum _{i=1}^n{i}^2={\displaystyle \frac{n(n+1)(2n+1)}{6}}}$	${\displaystyle \sum _{i=0}^n{r}^i={\displaystyle \frac{1-r^{n+1}}{1-r}}}$

1. $\lfloor {\log }_{2}h\rfloor$
2. $2h+1$
3. $2^{h+1}-1$
4. $(2h^3+3n^2+n)/6$
5. $h!$

The Wonder and Joy of Binary Trees

• We will spend a lot of time thinking about binary trees.
• Why????
• Question for you:
  • What is the maximum number of nodes in a binary tree of height $h$?
  • $N(h)\le 2^{h+1}-1$

Exercise #2

• What is the minimum height of a binary tree with $n$ nodes?
• $H(n)>=??$

1. $\lceil {\log }_{2}n\rceil$
2. $\lfloor {\log }_{2}n\rfloor$
3. $\lfloor \frac{n}{2}\rfloor$
4. $n$
5. $2^n$

Exercise #2

• What is the minimum height of a binary tree with $n$ nodes?
• $H(n)>=\lfloor {\log }_{2}n\rfloor$

Take Home Message

• In a linked list, we store a reference to one node, all other nodes are reachable by following $O(n)$ references.

• In a binary tree, we store a reference to one node, all other nodes are reachable by following $O({\log }_{2}n)$ references.

Pre-Order Traversal

static <E> void preorder(BinNode<E> rt) {
  if (rt == null) { return; } // Empty subtree - do nothing
  visit(rt);              // Process root node
  preorder(rt.left());    // Process all nodes in left
  preorder(rt.right());   // Process all nodes in right
}

In-Order Traversal

static <E> void inorder(BinNode<E> rt) {
  if (rt == null) { return; } // Empty subtree - do nothing
  inorder(rt.left());     // Process all nodes in left
  visit(rt);              // Process root node
  inorder(rt.right());    // Process all nodes in right
}

Post-Order Traversal

static <E> void postorder(BinNode<E> rt) {
  if (rt == null) { return; } // Empty subtree - do nothing
  postorder(rt.left());    // Process all nodes in left
  postorder(rt.right());   // Process all nodes in right
  visit(rt);               // Process root node
}

Another Traversal

Trace the following traversal method…

  public static <E> void mystery(BinNode<E> rt) {
    Queue<BinNode<E>> queue = new LQueue<>();

    queue.enqueue(rt);

    while (queue.length() > 0) {
      BinNode<E> cur = queue.dequeue();

      visit(cur);

      if (cur.left() != null) {
        queue.enqueue(cur.left());
      }
      if (cur.right() != null) {
        queue.enqueue(cur.right());
      }
    }

  }