求递归算法树 - ITeye问答

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求递归算法树3

id name parentId

TestDemo right6 = new TestDemo(6,"客户添加",5);
TestDemo right7 = new TestDemo(7,"客户删除",5);
TestDemo right8 = new TestDemo(8,"客户修改",5);
TestDemo right5 = new TestDemo(5,"客户管理",0);
TestDemo right10 = new TestDemo(10,"VIP客户添加",6);
TestDemo right11 = new TestDemo(11,"普通客户添加",6);
rightList.add(right6);
rightList.add(right7);
rightList.add(right8);
rightList.add(right10);
rightList.add(right5);
rightList.add(right11);
========================================================
最终要求效果：[拼接成以下字符串]
首先肯定是要判断有没有父类节点为0的有的话那么该条数据就为父节点

<sx:tree>
   <sx:treenode lable='客户管理'>
       <sx:treenode lable='客户添加'>
          <sx:treenode lable='VIP客户添加></sx:treenode>
          <sx:treenode lable='普通客户添加></sx:treenode>
       </sx:treenode>
       <sx:treenode lable='客户修改'></sx:treenode>
       <sx:treenode lable='客户删除'></sx:treenode>
   </sx:treenode>
</sx:tree>

客户添加修改删除的父节点都为5    上面的应该很清楚了
但是目前这个是2级如果3级甚至无限级怎么递归   我这只能递归2级如上图的但是考虑长远递归树的无限递归
求各位指点迷津了！小子不甚感激

Java 递归树

2012年4月22日 23:43

msdghs
2
0 0 22

2个答案按时间排序按投票排序

0 0

采纳的答案

public class TestDemo {
	private int id;
	private String name;
	private int parentId;
	private Set<TestDemo> child = new HashSet<TestDemo>();

	public String toXML(){
		StringBuffer xml = new StringBuffer();
		if(parentId==0){
			xml.append("<sx:tree>");
		}else{
			xml.append("<sx:treenode lable='"+name+"'>");
		}
		
		for(TestDemo td : child){
			xml.append(td.toXML());
		}
		
		if(parentId==0){
			xml.append("</sx:tree> ");
		}else{
			xml.append("</sx:treenode>");
		}
		return xml.toString();
	}
}

public static void main(String[] args) {
		List<TestDemo> rightList = new ArrayList<TestDemo>();
		TestDemo right6 = new TestDemo(6,"客户添加",5); 
		TestDemo right7 = new TestDemo(7,"客户删除",5); 
		TestDemo right8 = new TestDemo(8,"客户修改",5); 
		TestDemo right5 = new TestDemo(5,"客户管理",0); 
		TestDemo right10 = new TestDemo(10,"VIP客户添加",6); 
		TestDemo right11 = new TestDemo(11,"普通客户添加",6); 
		rightList.add(right6); 
		rightList.add(right7); 
		rightList.add(right8); 
		rightList.add(right10); 
		rightList.add(right5); 
		rightList.add(right11); 
		TestDemo root = getRoot(rightList);
		System.out.println(root.toXML());
	}

	public static TestDemo getRoot(List<TestDemo> list){
		TestDemo root = null;
		for(TestDemo td : list){
			if(td.getParentId()==0){
				root = td;
			}
		}
		initChild(root,list);
		return root;
	}
	public static void initChild(TestDemo tdemo,List<TestDemo> list){
		int parentId = tdemo.getId();
		for(TestDemo td : list){
			if(td.getParentId()==parentId){
				initChild(td,list);
				tdemo.getChild().add(td);
			}
		}
	}

2012年4月24日 17:45