Re: Last grandchild of a node in a tree [modified] - C# Discussion Boards

Last grandchild of a node in a tree

23-Oct-09 0:08

Because it would be pointless ? The args collection is passed in, and exists only inside this method, why would you have any desire to change it ?

Christian Graus

Driven to the arms of OSX by Vista.

Read my blog to find out how I've worked around bugs in Microsoft tools and frameworks.

Re: Why cant i do this?

dxlee23-Oct-09 5:00

dxlee

23-Oct-09 5:00

There is no element in the array args if you don't run the program with a command-line parameter. Therefore, args[0] does not exist. 0 is out of range.

devboycpp22-Oct-09 23:17

Re: Last grandchild of a node in a tree [modified]

22-Oct-09 23:17

In a recursive function for traversing a tree nodes , I want to invoke a function for the last child of a parent node( including all its grandchilds ). I have not found a technique for finding that node in the recursive function .

Any help ?
Thanks in advance.

dan!sh 23-Oct-09 0:27

Re: Last grandchild of a node in a tree [modified]

23-Oct-09 0:27

List<TreeNode> nodeList = new List<TreeNode>();
            private List<TreeNode> GetChildNodes(TreeNode treeNode)
            {
                  foreach (TreeNode oNode in treeNode.Nodes)
                  {
                        nodeList.Add(oNode);
                        if (oNode.Nodes.Count > 0)
                        {
                              GetChildNodes(oNode);
                        }
                        else {
                        // this is the last node
                        }
                  }
                  return nodeList;

            }

This would give you list of nodes in order as they would appear in expanded tree view.

It's not necessary to be so stupid, either, but people manage it. - Christian Graus, 2009 AD

modified on Friday, October 23, 2009 6:39 AM

devboycpp23-Oct-09 0:46

23-Oct-09 0:46

How to find the last grandchild of a parent node. All nodes have just one last grandchild.

in the following example node "node1_child3_child3" is the last grandchild of node1. Then how can I recognize that node in the recursive function ? What technique can I use ?

----node1
       node1_child1
       node1_child2
                node1_child2_child1
                node1_child2_child2
       node1_child3
                node1_child3_child1
                node1_child3_child2
                node1_child3_child3

modified on Friday, October 23, 2009 6:53 AM

dan!sh 23-Oct-09 1:11

23-Oct-09 1:11

If you need just the last node, change the return type of the method to treenode and use following code:

nodeList.Add(treeNode);
if (treeNode.Nodes.Count > 0) {
  treeNode = GetChildNodes(treeNode.Nodes[treeNode.Nodes.Count - 1]);
}
return treeNode;

If you need complete heirarchy, you the following:

nodeList.Add(treeNode);
if (treeNode.Nodes.Count > 0) {
   GetChildNodes(treeNode.Nodes[treeNode.Nodes.Count - 1]);
}
return nodeList;

It's not necessary to be so stupid, either, but people manage it. - Christian Graus, 2009 AD

devboycpp23-Oct-09 2:47

23-Oct-09 2:47

I need to find the node while traversing the tree in the recursive function.

dan!sh 23-Oct-09 3:55

23-Oct-09 3:55

Isn't that recursion? GetChildNodes calling itself?

It's not necessary to be so stupid, either, but people manage it. - Christian Graus, 2009 AD

devboycpp23-Oct-09 5:51

23-Oct-09 5:51

yes but it gives me no way to identify the last grandchild of a node .

ely_bob23-Oct-09 6:50

ely_bob

23-Oct-09 6:50

I used an addressing system for a similar problem(Drawing the tree)
here your address's woud be
0.0.0
0.1.0
0.2.0
0.2.1
0.2.2
0.3.0==node1_child3 Base node
0.3.1
0.3.2
0.3.3
**Would be 1.0.0... I think the answer is then obvious.

this pattern can then be applied to any depth of child...

devboycpp23-Oct-09 6:57

Re: Last grandchild of a node in a tree [modified]

23-Oct-09 6:57

Thanks for the reply. But I still don't get the approach ? can you elaborate please . The last grandchild here is 0.3.3.
How the traversing function would be then ?

ely_bob23-Oct-09 7:50

ely_bob

23-Oct-09 7:50

here is an approximation since I don't know your exact structure:

<pre>

class address
{
      list<int> token;

}

list<address> addresses = Tree.MapAddress();
int generation =0;
for(int i=0;i < addresses.Count-1;i++)
{
   if(addresses[i+1].token[generation] > addresses[i].token[generation])
   {      Console.Writeline("Your function result Here");}            }
}

</pre>

Now I had some wonky recursion like I would need to skip children(closed leaf) and step into/out of children so you can navigate (almost as fast) using the address and recursion... but it takes a LOT of code. this is the smallest patch I can think of for your situation.

also you can change the generation by swapping out the generation number so if you wanted the line between great grand children and grand children switch 0->1, and voila.

Poke tongue | ;-P

modified on Friday, October 23, 2009 2:01 PM

devboycpp23-Oct-09 8:17

23-Oct-09 8:17

ely_bob wrote:
int generation =0;
for(int i=0;i < addresses.Count-1;i++)
{
if(addresses[i+1].token[generation] > addresses[i].token[generation])
{ Console.Writeline("Your function result Here");} }
}

What happend to the "generation" variable ? It is set to 0 and remains the same. Are you sure it is right ?

ely_bob23-Oct-09 8:26

ely_bob

23-Oct-09 8:26

Yup.

That is a function stub, plop out that code and make a function with it:

 myfunction(int generation){
for(int i=0;i < addresses.Count-1;i++)
{
     if(addresses[i+1].token[generation] >addresses[i].token[generation])
     { Console.Writeline("Your function result Here");}

 }
}}

just in case you want to reuse that code for say.. color coding your decendants.. you could use generation as the switch value so that your generations are all displayed in different colors... for example.

basically i just did it that way to show: that number can be set as a variable so you can get more mileage out of your code.

How to check status file

Dedy Hartono22-Oct-09 22:59

Dedy Hartono

22-Oct-09 22:59

Hi all,

I am making a web application project. When the web application is run, the xml file will be always generated in a directory(app_data folder). The new xml file will overwrite the old one. I try to run with 2 browsers at once and one browser is successful and another one gets error and need to be refresh.
It is because the xml file is written by one application. So that I need to check the file status first when want to write. How to check the file status?Please include the code as it will very helpful for me. Thanks.

Regards,
Dedy

Christian Graus22-Oct-09 23:24

22-Oct-09 23:24

We have an ASP.NET forum.

The System.IO.File class probably does what you want. However, it sounds like what you're doing is stupid. Use a database to support multiple users, not an XML file

Christian Graus

Driven to the arms of OSX by Vista.

Read my blog to find out how I've worked around bugs in Microsoft tools and frameworks.

dan!sh 22-Oct-09 23:37

22-Oct-09 23:37

Only way to check this is to try open the file. If you are using File.Open, one of the overload takes in FileShare as a parameter. That will allow muliple access. Although, not sure about what will happen when you simultaneously write something in the file from two sources.

It's not necessary to be so stupid, either, but people manage it. - Christian Graus, 2009 AD

Christian Graus22-Oct-09 23:39

22-Oct-09 23:39

d@nish wrote:
Although, not sure about what will happen when you simultaneously write something in the file from two sources.

The correct term is 'meltdown'. Well, it's 'race condition', but I think 'meltdown' might better illustrate the end result.

Christian Graus

Driven to the arms of OSX by Vista.

Read my blog to find out how I've worked around bugs in Microsoft tools and frameworks.

dan!sh 22-Oct-09 23:58

22-Oct-09 23:58

OK. Thanks. Smile | :)

But I don't think OP would be using this since it is an XML and .Net does provide specific classes for that.

Btw, I never seem to understand your timezone.

It's not necessary to be so stupid, either, but people manage it. - Christian Graus, 2009 AD

Christian Graus23-Oct-09 0:09