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. $2^h+h$
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 depth 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 were reachable by following $O({\log }_{2}n)$ references.

Traversals

Traversals

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());
      }
    }

  }