2.2.1.1. Binary Trees¶

A binary tree is made up of a finite set of nodes that is either empty or consists of a node called the root together with two binary trees, called the left and right subtrees, which are disjoint from each other and from the root.

Notation: Node, children, edge, parent, ancestor, descendant, path, depth, height, level, leaf node, internal node, subtree.

2.2.1.2. A Recursive Data Structure¶

2.2.1.3. Binary Tree Node Class¶

interface BinNode { // Binary tree node ADT
  // Get and set the element value
  public Object value();
  public void setValue(Object v);

  // return the children
  public BinNode left();
  public BinNode right();

  // return TRUE if a leaf node, FALSE otherwise
  public boolean isLeaf();
}

2.2.1.4. Question¶

• Write a recursive function named count that, given the root to a binary tree, returns a count of the number of nodes in the tree. Function count should have the following prototype:

  int count(BinNode root)
  

2.2.1.5. Traversals¶

• Any process for visiting the nodes in some order is called a traversal.

• Any traversal that lists every node in the tree exactly once is called an enumeration of the tree’s nodes.

• Preorder traversal: Visit each node before visiting its children.

• Postorder traversal: Visit each node after visiting its children.

• Inorder traversal: Visit the left subtree, then the node, then the right subtree.

2.2.1.6. Preorder Traversal (1)¶

static void preorder(BinNode 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
}

2.2.1.7. Preorder Traversal (2)¶

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2.2.1.8. How not to write a traversal¶

// This is a bad idea
static void preorder2(BinNode rt) {
  visit(rt);
  if (rt.left() != null) preorder2(rt.left());
  if (rt.right() != null) preorder2(rt.right());
}

Problems:

This has a major bug

It puts the focus in the wrong place: Should focus on the current node, not the children. This version is therefore more complicated.

2.2.1.9. Recursion Examples¶

static int count(BinNode rt) {
  if (rt == null) return 0;  // Nothing to count
  return 1 + count(rt.left()) + count(rt.right());
}

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