今天遇到一个求树形结构中所有节点的值求合的问题,TreeList自身提供了类似的功能,可是它的表现形式并不是我想要的,我需要将子结点的合放在父节点的值中。折腾了几下。还弄不出来,我是不喜欢接受人家东西的人。除非是自己实现不了。那是能力问题,自己想了一下。有些头绪,干脆自己来写个过程实现算了。刚开始想着用递归。说来惭愧。学编程这么久。对这个东西还是模模糊糊的印象。搞了两下。竟然不行。上网一问大虾们。被骂个半死。偶是菜鸟中的菜鸟,连二叉树的问题都搞不定。说的我当时就想去跳楼。没有办法。能力有限啊。最后想到了将树形数据转化为线型数据来处理,这样子就发挥了搜索和排序的威力。最后用Hashtable实现。到于线型数据,我用到了TreeListOperation中的Execute过程来将所有的节点加到一个Hashtable中。转换过程代码如下:
//对树形数据求和用到数据结构
public struct TreeSumData
{
public int Level;//深度
public double Value;//值
public int ParentID;//父节点ID
public int ID;//自身ID
public bool IsParent;//是否为父节点
}
当得到一个线型的数据列表和该树形数据集合的最大深度后,我们就可以正式开始进行计算了。菜鸟我不会二叉树。也不会什么算法。就只好一层一层的算了。中间有大量的拆箱和装箱操作,在2005下采用List泛型实现效率应该会高一些。
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using System;
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using System.Collections ;
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using DevExpress.XtraTreeList;
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using DevExpress.XtraTreeList.Columns ;
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using DevExpress.XtraTreeList.Nodes ;
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using DevExpress.XtraTreeList.Nodes.Operations ;
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namespace NskProject
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{
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/// <summary>
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///求树形数据之和,辅助类,将数据取到Hashtable。
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/// </summary>
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public class GetSumItemList:TreeListOperation
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{
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Hashtable _ValueList=new Hashtable();
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int _MaxLevel=0;
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public override void Execute(TreeListNode node)
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{
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TreeSumData SumData=new TreeSumData();
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SumData.Level =node.Level;
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if(SumData.Level>_MaxLevel)
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{
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_MaxLevel=SumData.Level ;
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}
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SumData.ParentID =node.ParentNode!=null?node.ParentNode.Id+1:0;
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SumData.ID =node.Id+1;
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string text=node.GetDisplayText(ValueColumn);
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double value=0;
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if(Double.TryParse(text,System.Globalization.NumberStyles.Number,null,out value))
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{
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SumData.Value =value;
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}
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else
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{
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SumData.Value =0;
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}
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SumData.IsParent =node.HasChildren ?true:false;
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ValueList.Add(node.Id+1 ,SumData);
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}
39![](https://www.cnblogs.com/Images/OutliningIndicators/InBlock.gif)
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TreeListColumn _ValueColumn=null;
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public TreeListColumn ValueColumn
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{
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get{return _ValueColumn;}
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set{_ValueColumn=value;}
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}
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public Hashtable ValueList
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{
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get{return _ValueList;}
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}
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public int MaxLevel
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{
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get{return _MaxLevel;}
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}
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}
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}
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其中存入到Hashtable中的Struct是这样的:![](https://www.cnblogs.com/Images/OutliningIndicators/None.gif)
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//对树形数据求和用到数据结构
public struct TreeSumData
{
public int Level;//深度
public double Value;//值
public int ParentID;//父节点ID
public int ID;//自身ID
public bool IsParent;//是否为父节点
}
当得到一个线型的数据列表和该树形数据集合的最大深度后,我们就可以正式开始进行计算了。菜鸟我不会二叉树。也不会什么算法。就只好一层一层的算了。中间有大量的拆箱和装箱操作,在2005下采用List泛型实现效率应该会高一些。
1
using System;
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using System.Collections ;
3![](https://www.cnblogs.com/Images/OutliningIndicators/None.gif)
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namespace NskProject
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{
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/// <summary>
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/// 自顶向上计算各个节点的值
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/// </summary>
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public class SumTreeDataInHashtable
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{
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Hashtable Value=new Hashtable();
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public SumTreeDataInHashtable(Hashtable value)
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{
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Value=value;
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}
16![](https://www.cnblogs.com/Images/OutliningIndicators/InBlock.gif)
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public void SumAll()
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{
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for(int i=MaxLevel-1;i>=0;i--)
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{
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GetParentList(i);
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}
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}
24![](https://www.cnblogs.com/Images/OutliningIndicators/InBlock.gif)
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//最大深度
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int _MaxLevel=0;
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public int MaxLevel
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{
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get{return _MaxLevel;}
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set{_MaxLevel=value;}
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}
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public Hashtable Result
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{
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get{return Value;}//结果输出
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}
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//单层的计算
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private void GetParentList(int Level)
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{
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ArrayList ID;
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ID=GetParentID(Level);
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for(int n=0;n<ID.Count;n++)
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{
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int CurrentParentID=Convert.ToInt32(ID[n]);
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ArrayList Current=new ArrayList();
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foreach(object value in Value.Values)
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{
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TreeSumData Data=(TreeSumData)value;
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if(Data.ParentID==CurrentParentID)
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{
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Current.Add(Data.Value);
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}
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else if(Data.ID==CurrentParentID)
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{
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Current.Add(Data.Value );
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}
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}
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double SumValue=SumArrayList(Current);
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TreeSumData tmp=(TreeSumData)Value[CurrentParentID];
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tmp.Value =SumValue;
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Value[CurrentParentID]=tmp;
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}
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}
63![](https://www.cnblogs.com/Images/OutliningIndicators/InBlock.gif)
64![](https://www.cnblogs.com/Images/OutliningIndicators/InBlock.gif)
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/// <summary>
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/// 取得上层父节点有的数量
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/// </summary>
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/// <param name="Level"></param>
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/// <returns></returns>
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private ArrayList GetParentID(int Level)
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{
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ArrayList Res=new ArrayList();
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foreach(object value in Value.Values)
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{
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TreeSumData Data=(TreeSumData)value;
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if(Data.Level ==Level)
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{
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if(!Res.Contains(Data.ID))
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{
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Res.Add(Data.ID);
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}
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}
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}
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return Res;
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}
87![](https://www.cnblogs.com/Images/OutliningIndicators/InBlock.gif)
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//计算一个ArrayList中所有值的和
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private double SumArrayList(ArrayList Source)
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{
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double Res=0;
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for(int i=0;i<Source.Count;i++)
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{
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double t=Convert.ToDouble(Source[i]);
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Res+=t;
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}
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return Res;
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}
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}
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}
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