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>
 * 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>
 * 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:

[quote]

A B B1

A B B4

A C C2

A C C4

A C C5

[/quote]