java 广度优先

就是按照正常顺序把tree打出来，假设有个tree是
       1
     /   \
    2     3
   / \   / \
  4   5  6  7
打印出来就是 1，2，3，4，5，6，7；
这个就是宽度优先

package org.iaiai.suanfa;

import java.util.LinkedList;

/**
 * 
 * <p>
 * Title: Node.java
 * </p>
 * <p>
 * E-Mail: [e-mail]
 * </p>
 * <p>
 * QQ: 176291935
 * </p>
 * <p>
 * Http: iaiai.iteye.com
 * </p>
 * <p>
 * Create time: 2011-8-5
 * </p>
 * 
 * @author 丸子
 * @version 0.0.1
 */
public class Node {

	public int level = 0;
	public String path = " ";
	public LinkedList<Node> children = new LinkedList<Node>();
	public String value = " ";
	public String name = " ";

	/**
	 * 
	 * @param val
	 *            第几级
	 * @param name
	 *            名子
	 * @param parent
	 *            父级
	 */
	public Node(String val, String name, Node parent) {
		this.value = val;
		this.name = name;
		if (parent != null) {
			parent.addChild(this);
			path = parent.path + "   " + name;
		} else
			path = name;
	}

	public void addChild(Node child) {
		children.add(child);
	}

}

package org.iaiai.suanfa;

import java.util.LinkedList;
import java.util.List;

/**
 * 
 * <p>
 * Title: Main.java
 * </p>
 * <p>
 * E-Mail: [e-mail]
 * </p>
 * <p>
 * QQ: 176291935
 * </p>
 * <p>
 * Http: iaiai.iteye.com
 * </p>
 * <p>
 * Create time: 2011-8-5
 * </p>
 * 
 * @author 丸子
 * @version 0.0.1
 */
public class Main {

	public static void main(String[] args) throws Exception {
		int n = 3;
		String[] str = { "1", "2", "3" };
		Node A = new Node("1", "A", null);
		Node B = new Node("2", "B", A);
		Node C = new Node("2", "C", A);
		new Node("3", "B1", B);
		new Node("x", "B2", B); // x为不显示
		new Node("x", "B3", B);
		new Node("3", "B4", B);
		new Node("x", "C1", C);
		new Node("3", "C2", C);
		new Node("x", "C3", C);
		new Node("3", "C4", C);
		new Node("3", "C5", C);

		List<Node> queue = new LinkedList<Node>();
		if (A.value.equals(str[A.level]))
			queue.add(A);
		while (queue.size() > 0) {
			Node cur = queue.remove(0);
			LinkedList<Node> children = cur.children;
			for (int i = 0; i < children.size(); i++) {
				if (children.get(i).value.equals(str[cur.level + 1])) {
					if (cur.level + 1 == n - 1)
						System.out.println(children.get(i).path);
					else {
						children.get(i).level = cur.level + 1;
						queue.add(children.get(i));
					}
				}
			}
		}
	}

}

输出:

引用

A   B   B1
A   B   B4
A   C   C2
A   C   C4
A   C   C5