Somehow the 2-ary trees are the most common. Really common, as it turns out.

A binary tree is just a k-ary tree with k = 2. Because k is only 2,

- Binary trees are a bit simpler and easier to understand than trees with a large or unbounded number of children
- There are special traversals besides the usual breadth-first and depth-first traversals
- It is fun (or at least a valuable brain exercise) to generate the formula for the number of distinct binary tree shapes for a given number of nodes
- Binary tree nodes have an elegant linked representation with "left" and "right" subtrees
- Binary trees form the basis for efficient representations of sets, dictionaries, and priority queues

Like all oriented trees, breadth-first and depth-first traversals exist for binary trees, but there are others:

**Preorder**: C F G A D E H B I J K**Inorder**: G F D A E C I B K J H**Postorder**: G D E A F I K J B H C

Three simple rules:

- Write the empty tree as:
`()`

- Write a non empty tree as:
`(`

*leftSubtree root rightSubtree*) - An empty subtree can be, and usually is, omitted.

So for example:

- The tree with one element A is:
`(() A ())`

, or just`(A)`

- The tree with A as the root and B as the left child is:
`((() B ()) A ())`

, or just`((B) A)`

- The tree with A as the root and B as the right child is:
`(() A (() B ()))`

, or just`(A (B))`

- The tree in the previous section is:
`(((G) F (D (A) E)) C (((I) B ((K) J)) H))`

There are five distinct shapes of binary trees with three nodes:

But how many are there for n nodes?

Let C(n) be the number of distinct binary trees with n nodes. This is equal to the number of trees that have a root, a left subtree with j nodes, and a right subtree of (n-1)-j nodes, for each j. That is,

C(n) = C(0)C(n-1) + C(1)C(n-2) + ... + C(n-1)C(0)

which is

The first few terms:

C(0) = 1 C(1) = C(0)C(0) = 1 C(2) = C(0)C(1) + C(1)C(0) = 2 C(3) = C(0)C(2) + C(1)C(1) + C(2)C(0) = 5 C(4) = C(0)C(3) + C(1)C(2) + C(2)C(1) + C(3)C(0) = 14

You can prove

Here's the number of 8-node binary trees:

1 16! 16×15×14×13×12×11×10 C(8) = - × ---- = -------------------- = 13×11×10 = 1430 9 8!8! 8×7×6×5×4×3×2×1

Also see Wikipedia's article on the Catalan Numbers.

As is common with trees (though quite unlike lists), node classes are visible to clients. Here is a simple binary tree node interface, again using the visitor pattern for traversals.

BinaryTreeNode.java

/** * A simple interface for binary trees. An empty binary tree is represented with * the value null; a non-empty tree by its root node. */ public interface BinaryTreeNode<E> { /** * Returns the data stored in this node. */ E getData(); /** * Modifies the data stored in this node. */ void setData(E data); /** * Returns the parent of this node, or null if this node is a root. */ BinaryTreeNode<E> getParent(); /** * Returns the left child of this node, or null if it does not have one. */ BinaryTreeNode<E> getLeft(); /** * Removes child from its current parent and inserts it as the left child of * this node. If this node already has a left child it is removed. * * @exception IllegalArgumentException if the child is an ancestor of this node, * since that would make a cycle in the * tree. */ void setLeft(BinaryTreeNode<E> child); /** * Returns the right child of this node, or null if it does not have one. */ BinaryTreeNode<E> getRight(); /** * Removes child from its current parent and inserts it as the right child of * this node. If this node already has a right child it is removed. * * @exception IllegalArgumentException if the child is an ancestor of this node, * since that would make a cycle in the * tree. */ void setRight(BinaryTreeNode<E> child); /** * Removes this node, and all its descendants, from whatever tree it is in. Does * nothing if this node is a root. */ void removeFromParent(); /** * Visits the nodes in this tree in preorder. */ void traversePreorder(Visitor visitor); /** * Visits the nodes in this tree in postorder. */ void traversePostorder(Visitor visitor); /** * Visits the nodes in this tree in inorder. */ void traverseInorder(Visitor visitor); /** * Simple visitor interface. */ public interface Visitor { <E> void visit(BinaryTreeNode<E> node); } }

We can implement this interface with links

LinkedBinaryTreeNode.java

/** * An implementation of the BinaryTreeNode interface in which each node stores * direct links to its left child, its right child, and its parent. * * <p> * LinkedBinaryTreeNode objects are pretty mean: if one tries to mix them up * with different kinds of binary tree nodes, and exception may be thrown. * </p> */ public class LinkedBinaryTreeNode<E> implements BinaryTreeNode<E> { protected E data; protected LinkedBinaryTreeNode<E> parent; protected LinkedBinaryTreeNode<E> left; protected LinkedBinaryTreeNode<E> right; /** * Constructs a node as the root of its own one-element tree. This is the only * public constructor. The only trees that clients can make directly are simple * one-element trees. */ public LinkedBinaryTreeNode(E data) { this.data = data; } /** * Returns the data stored in this node. */ public E getData() { return data; } /** * Modifies the data stored in this node. */ public void setData(E data) { this.data = data; } /** * Returns the parent of this node, or null if this node is a root. */ public BinaryTreeNode<E> getParent() { return parent; } /** * Returns the left child of this node, or null if it does not have one. */ public BinaryTreeNode<E> getLeft() { return left; } /** * Removes child from its current parent and inserts it as the left child of * this node. If this node already has a left child it is removed. * * @exception IllegalArgumentException if the child is an ancestor of this node, * since that would make a cycle in the * tree. */ public void setLeft(BinaryTreeNode<E> child) { // Ensure the child is not an ancestor. for (LinkedBinaryTreeNode<E> n = this; n != null; n = n.parent) { if (n == child) { throw new IllegalArgumentException(); } } // Ensure that the child is an instance of LinkedBinaryTreeNode. LinkedBinaryTreeNode<E> childNode = (LinkedBinaryTreeNode<E>) child; // Break old links, then reconnect properly. if (this.left != null) { left.parent = null; } if (childNode != null) { childNode.removeFromParent(); childNode.parent = this; } this.left = childNode; } /** * Returns the right child of this node, or null if it does not have one. */ public BinaryTreeNode<E> getRight() { return right; } /** * Removes child from its current parent and inserts it as the right child of * this node. If this node already has a right child it is removed. * * @exception IllegalArgumentException if the child is an ancestor of this node, * since that would make a cycle in the * tree. */ public void setRight(BinaryTreeNode<E> child) { // Ensure the child is not an ancestor. for (LinkedBinaryTreeNode<E> n = this; n != null; n = n.parent) { if (n == child) { throw new IllegalArgumentException(); } } // Ensure that the child is an instance of LinkedBinaryTreeNode. LinkedBinaryTreeNode<E> childNode = (LinkedBinaryTreeNode<E>) child; // Break old links, then reconnect properly. if (right != null) { right.parent = null; } if (childNode != null) { childNode.removeFromParent(); childNode.parent = this; } this.right = childNode; } /** * Removes this node, and all its descendants, from whatever tree it is in. Does * nothing if this node is a root. */ public void removeFromParent() { if (parent != null) { if (parent.left == this) { parent.left = null; } else if (parent.right == this) { parent.right = null; } this.parent = null; } } /** * Visits the nodes in this tree in preorder. */ public void traversePreorder(BinaryTreeNode.Visitor visitor) { visitor.visit(this); if (left != null) left.traversePreorder(visitor); if (right != null) right.traversePreorder(visitor); } /** * Visits the nodes in this tree in postorder. */ public void traversePostorder(Visitor visitor) { if (left != null) left.traversePostorder(visitor); if (right != null) right.traversePostorder(visitor); visitor.visit(this); } /** * Visits the nodes in this tree in inorder. */ public void traverseInorder(Visitor visitor) { if (left != null) left.traverseInorder(visitor); visitor.visit(this); if (right != null) right.traverseInorder(visitor); } }

For fun, here is a panel class that can draw a binary tree.

BinaryTreePanel.java

import java.awt.Color; import java.awt.FontMetrics; import java.awt.Graphics; import java.awt.Point; import java.awt.Rectangle; import java.lang.reflect.Field; import java.util.HashMap; import java.util.Map; import javax.swing.JPanel; /** * A panel that maintains a picture of a binary tree. */ public class BinaryTreePanel extends JPanel { private BinaryTreeNode<?> tree; private int gridwidth; private int gridheight; /** * Stores the pixel values for each node in the tree. */ private Map<BinaryTreeNode<?>, Point> coordinates = new HashMap<BinaryTreeNode<?>, Point>(); /** * Constructs a panel, saving the tree and drawing parameters. */ public BinaryTreePanel(BinaryTreeNode<?> tree, int gridwidth, int gridheight) { this.tree = tree; this.gridwidth = gridwidth; this.gridheight = gridheight; } /** * Changes the tree rendered by this panel. */ public void setTree(BinaryTreeNode<?> root) { tree = root; repaint(); } /** * Draws the tree in the panel. First it computes the coordinates of all the * nodes with an inorder traversal, then draws them with a postorder traversal. */ public void paintComponent(final Graphics g) { super.paintComponent(g); if (tree == null) { return; } tree.traverseInorder(new BinaryTreeNode.Visitor() { private int x = gridwidth; public void visit(BinaryTreeNode node) { coordinates.put(node, new Point(x, gridheight * (depth(node) + 1))); x += gridwidth; } }); tree.traversePostorder(new BinaryTreeNode.Visitor() { public void visit(BinaryTreeNode node) { String data = node.getData().toString(); Point center = (Point) coordinates.get(node); if (node.getParent() != null) { Point parentPoint = (Point) coordinates.get(node.getParent()); g.setColor(Color.black); g.drawLine(center.x, center.y, parentPoint.x, parentPoint.y); } FontMetrics fm = g.getFontMetrics(); Rectangle r = fm.getStringBounds(data, g).getBounds(); r.setLocation(center.x - r.width / 2, center.y - r.height / 2); Color color = getNodeColor(node); Color textColor = (color.getRed() + color.getBlue() + color.getGreen() < 382) ? Color.white : Color.black; g.setColor(color); g.fillRect(r.x - 2, r.y - 2, r.width + 4, r.height + 4); g.setColor(textColor); g.drawString(data, r.x, r.y + r.height); } }); } /** * Returns a color for the node. If the node is of a class with a field called * "color", and that field currently contains a non-null value, then that value * is returned. Otherwise a default color of yellow is returned. */ Color getNodeColor(BinaryTreeNode<?> node) { try { Field field = node.getClass().getDeclaredField("color"); return (Color) field.get(node); } catch (Exception e) { return Color.yellow; } } private int depth(BinaryTreeNode<?> node) { return (node.getParent() == null) ? 0 : 1 + depth(node.getParent()); } }