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  • Reduced Tree View in NetBeans IDE 7.2

    - by Geertjan
    Right-click within the Projects window in NetBeans IDE 7.2 and from the "View Java Packages As" menu, you can now choose "Reduced Tree".I never really understood the difference between "Reduced Tree" and the already existing "Tree". But it makes sense when you see it. Here's Reduced Tree view: And here's Tree view, where you can see that the "actions" and "nodes" packages above each have their own top level package nodes, which takes up more space than the above: What's cool is that your selected package view is persisted across restarts of the IDE. To be complete, here's the List view, which is the third option you have in the "View Java Packages As" menu: Seems to me like the new Reduced Tree view combines the best of the Tree view with the best of the List view! Related issue: http://netbeans.org/bugzilla/show_bug.cgi?id=53192

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  • check if a tree is a binary search tree

    - by TimeToCodeTheRoad
    I have written the following code to check if a tree is a Binary search tree. Please help me check the code: Pair p{ boolean isTrue; int min; int max; } public boo lean isBst(BNode v){ return isBST1(v).isTrue; } public Pair isBST1(BNode v){ if(v==null) return new Pair(true, INTEGER.MIN,INTEGER.MAX); if(v.left==null && v.right==null) return new Pair(true, v.data, v.data); Pair pLeft=isBST1(v.left); Pair pRight=isBST1(v.right); boolean check=pLeft.max<v.data && v.data<= pRight.min; Pair p=new Pair(); p.isTrue=check&&pLeft.isTrue&&pRight.isTrue; p.min=pLeft.min; p.max=pRight.max; return p; } Note: This function checks if a tree is a binary search tree

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  • how does NTFS actually work with B-tree ?

    - by bakra
    To improve performance, NTFS directories use a special data management structure called a B-tree. "B-tree" concept here refers to a "tree of storage units" that hold the contents of an individual directory. What I don't understand is where on the disk is this tree stored? Its surely not created every-time we reboot...that would take lots of time. and since its a tree(dynamic Data structure) unlike arrays it will grow. so space needs to be allocated every-time it grows. so how is this "dynamic meta-data" stored ?

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  • Accessing selected node of richfaces tree from Javascript

    - by kazanaki
    Hello This should be a very simple question. I have a richfaces tree that is rendered using JSF. When the user clicks on a node I want a javascript function to run. Nothing more nothing less. No redirects, no re-submit, no-rerender, no Ajax. Just plain old Javascript. I have seen the onselected attribute of the tree and it indeed fires a Javascript method. But of course I want to know which node was clicked. Here is what I have so far <head> <script type="text/javascript"> function documentClicked(nodeRef) { alert("Node is "+nodeRef); } </script> </head> <rich:tree switchType="client" value="#{ajaxDocumentTree.rootNode}" var="document" onselected="documentClicked()" > <rich:treeNode iconLeaf="../images/tree/doc.gif" icon="../images/tree/doc.gif"> <h:outputText value="#{doc.friendlyName}" /> </rich:treeNode> But this does not work because nodeRef is undefined. I expected that the first argument of the callback would be the selected node but this is not the case. So the question is this: How do I fire a Javascript function with the selected node from a richfaces tree?

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  • JQGrid tree - passing additional parameters when tree is expanded

    - by PHP thinker
    I have a JQGRid tree. It loads data click by click, not all at once. Typically, JQGRid passes 4 standard tree parameters with each call - row (level, parent, is leaf, is expanded). How can I pass more parameters that I will take from the row being expanded? E.g. data from Name column should be passed in AJAX call too. There doesn't seem to be OnExpand event or similar.

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  • k-d tree implementation [closed]

    - by user466441
    when i run my code and debugged,i got this error - this 0x00093584 {_Myproxy=0x00000000 _Mynextiter=0x00000000 } std::_Iterator_base12 * const - _Myproxy 0x00000000 {_Mycont=??? _Myfirstiter=??? } std::_Container_proxy * _Mycont CXX0017: Error: symbol "" not found _Myfirstiter CXX0030: Error: expression cannot be evaluated + _Mynextiter 0x00000000 {_Myproxy=??? _Mynextiter=??? } std::_Iterator_base12 * but i dont know what does it means,code is this #include<iostream> #include<vector> #include<algorithm> using namespace std; struct point { float x,y; }; vector<point>pointleft(4); vector<point>pointright(4); //we are going to implement two comparison function for x and y coordinates,we need it in calculation of median (we should sort vector //by x or y according to depth informaton,is depth even or odd. bool sortby_X(point &a,point &b) { return a.x<b.x; } bool sortby_Y(point &a,point &b) { return a.y<b.y; } //so i am going to implement to median finding algorithm,one for finding median by x and another find median by y point medianx(vector<point>points) { point temp; sort(points.begin(),points.end(),sortby_X); temp=points[(points.size()/2)]; return temp; } point mediany(vector<point>points) { point temp; sort(points.begin(),points.end(),sortby_Y); temp=points[(points.size()/2)]; return temp; } //now construct basic tree structure struct Tree { float x,y; Tree(point a) { x=a.x; y=a.y; } Tree *left; Tree *right; }; Tree * build_kd( Tree *root,vector<point>points,int depth) { point temp; if(points.size()==1)// that point is as a leaf { if(root==NULL) root=new Tree(points[0]); return root; } if(depth%2==0) { temp=medianx(points); root=new Tree(temp); for(int i=0;i<points.size();i++) { if (points[i].x<temp.x) pointleft[i]=points[i]; else pointright[i]=points[i]; } } else { temp=mediany(points); root=new Tree(temp); for(int i=0;i<points.size();i++) { if(points[i].y<temp.y) pointleft[i]=points[i]; else pointright[i]=points[i]; } } return build_kd(root->left,pointleft,depth+1); return build_kd(root->right,pointright,depth+1); } void print(Tree *root) { while(root!=NULL) { cout<<root->x<<" " <<root->y; print(root->left); print(root->right); } } int main() { int depth=0; Tree *root=NULL; vector<point>points(4); float x,y; int n=4; for(int i=0;i<n;i++) { cin>>x>>y; points[i].x=x; points[i].y=y; } root=build_kd(root,points,depth); print(root); return 0; } i am trying ti implement in c++ this pseudo code tuple function build_kd_tree(int depth, set points): if points contains only one point: return that point as a leaf. if depth is even: Calculate the median x-value. Create a set of points (pointsLeft) that have x-values less than the median. Create a set of points (pointsRight) that have x-values greater than or equal to the median. else: Calculate the median y-value. Create a set of points (pointsLeft) that have y-values less than the median. Create a set of points (pointsRight) that have y-values greater than or equal to the median. treeLeft = build_kd_tree(depth + 1, pointsLeft) treeRight = build_kd_tree(depth + 1, pointsRight) return(median, treeLeft, treeRight) please help me what this error means?

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  • Extjs: Tree, Selecting node after creating the tree

    - by Natkeeran
    I have a simple TreePanel. I would like to select a particular node upon loading it. The nodes are from a remote file (json). The tree is loading as expected. However, the node is not being selected. Firebug shows node as undefined. This perhaps because of the async property. But, I an unable to configure this other wise, or specify the node be selected. Any suggestions welcomed, and thank you. LeftMenuTree = new Ext.tree.TreePanel({ renderTo: 'TreeMenu', collapsible: false, height: 450, border: false, userArrows: true, animate: true, autoScroll: true, id: 'testtest', dataUrl: fileName, root: { nodeType: 'async', iconCls:'home-icon', expanded:true, text: rootText }, listeners: { "click": { fn: onPoseClick, scope: this } }, "afterrender": { fn: setNode, scope: this } }); function setNode(){ alert (SelectedNode); if (SelectedNode == "Orders"){ var treepanel = Ext.getCmp('testtest'); var node = treepanel.getNodeById("PendingItems"); node.select(); } }

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  • SVN project folder tree structure vs production folder tree structure

    - by Marco Demaio
    While developing a PHP+JS web application we always try to separate big blocks of code into small modules/components, in order to make these last ones as much reusable as possible in other applications. Let's say we now have: the EcommerceApp (an ecommerce main application) a Server-file-mgr component (a component to view/manage file on server) a Mylib (a library of useful functions) a MailistApp (another main application to handle mail lists) ... EcommerceApp needs both Server-file-mgr component and Mylib to work Server-file-mgr needs Mylib to work MaillistApp needs both Server-file-mgr component and Mylib to work too. My idea is to simply structure the SVN project folder tree putting everything at the same level: trunk/EcommerceApp trunk/Server-file-mgr trunk/Mylib trunk/MaillistApp But in real life to make these apps to work the folder tree structure must be the following: EcommerceApp |_ Mylib |_ Server-file-mgr MaillistApp |_ Mylib |_ Server-file-mgr I mean Mylib and Server-file-mgr needs to be inside the EcommerceApp/MaillistApp folder. How would you then structure the SVN folder, as I did or in a different/better/smarter way???

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  • Insert element into a tree from a list in Standard ML

    - by vichet
    I have just started to learn SML on my own and get stuck with a question from the tutorial. Let say I have: tree data type datatype node of (tree*int*tree) | null insert function fun insert (newItem, null) = node (null, newItem, null) | insert (newItem, node (left, oldItem, right)) = if (newItem <= oldItem) then node (insert(newItem,left),oldItem, right) else node (left, oldItem, insert(newItem, right) an integer list val intList = [19,23,21,100,2]; my question is how can I add write a function to loop through each element in the list and add to a tree? Your answer is really appreciated.

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  • Maximum depth of a B-tree

    - by Phenom
    How do you figure out the maximum depth of a B-tree? Say you had a B-tree of order 1625, meaning each node has 1625 pointers and 1624 elements. What is the maximum depth of the tree if it contains 85,000,000 keys?

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  • Sorting by custom field and fetching whole tree from DB

    - by Niaxon
    Hello everyone, I am trying to do file browser in a tree form and have a problem to sort it somehow. I use PHP and MySQL for that. I've created mixed (nested set + adjacency) table 'element' with the following fields: element_id, left_key, right_key, level, parent_id, element_name, element_type (enum: 'folder','file'), element_size. Let's not discuss right now that it is better to move information about element (name, type, size) into other table. Function to scan specified directory and fill table work correctly. Noteworthy, i am adding elements to tree in specific order: folders first and then files. After that i can easily fetch and display whole table on the page using simple query: SELECT * FROM element WHERE 1=1 ORDER BY left_key With the result of that query and another function i can generate correct html code (<ul><li>... and so on). to display tree. Now back to the question (finally, huh?). I am struggling to add sorting functionality. For example i want to order my result by size. Here i need to keep in my mind whole hierarchy of tree and rule: folders first, files later. I believe i can do that by generating in PHP recursive query: SELECT * FROM element WHERE parent_id = {$parentId} ORDER BY element_type (so folders would be first), size (or name for example) asc/desc After that for each result which has type = 'folder' i will send another query to get it's content. Also it's possible to fetch whole tree by left_key and after that sort it in PHP as array but i guess that would be worse :) I wonder if there is better and more efficient way to do such a thing?

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  • Family Tree :- myheritage.com

    - by Nitesh Panchal
    Hello, The other day i just accidently visited the site myheritage.com. I was just wondering, how they must have created one? Can anybody tell me what can be their database design? and if possible, algorithm that we can use to generate such a tree? Generating simple binary tree is very easy using recursion. But if you have a look at the site(if you have time please make account on it and add few nodes to feel) when we add son to a father, it's mother is automatically added(if you don't add explicitly). Mother's family tree is also generated side by side and many such fancy things are happening. In a simple binary tree we have a root node and then many nodes below it. Thus we cannot show wife and husband in the tree and then show a line from wife and husband to child. In spare time, can anybody discuss what can be it's database design and the recursive algorithm that we can follow to generate it? I hope i am not asking too much from you :).

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  • How to Populate a 'Tree' structure 'Declaratively'

    - by mackenir
    I want to define a 'node' class/struct and then declare a tree of these nodes in code in such a way that the way the code is formatted reflects the tree structure, and there's not 'too much' boiler plate in the way. Note that this isn't a question about data structures, but rather about what features of C++ I could use to arrive at a similar style of declarative code to the example below. Possibly with C++0X this would be easier as it has more capabilities in the area of constructing objects and collections, but I'm using Visual Studio 2008. Example tree node type: struct node { string name; node* children; node(const char* name, node* children); node(const char* name); }; What I want to do: Declare a tree so its structure is reflected in the source code node root = node("foo", [ node("child1"), node("child2", [ node("grand_child1"), node("grand_child2"), node("grand_child3" ]), node("child3") ]); NB: what I don't want to do: Declare a whole bunch of temporary objects/colls and construct the tree 'backwards' node grandkids[] = node[3] { node("grand_child1"), node("grand_child2"), node("grand_child3" }; node kids[] = node[3] { node("child1"), node("child2", grandkids) node("child3") }; node root = node("foo", kids);

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  • unique substrings using suffix tree

    - by user1708762
    For a given string S of length n- Optimal algorithm for finding all unique substrings of S can't be less than O(n^2). So, the best algorithm will give us the complexity of O(n^2). As per what I have read, this can be implemented by creating suffix tree for S. The suffix tree for S can be created in O(n) time. Now, my question is- How can we use the suffix tree for S to get all the unique substrings of S in O(n^2)?

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  • Single-port 2600 router with 2900XL switch

    - by Slava Maslennikov
    I have a setup, where the single port 2600 router is in port 0/2 in the switch, outside network is on port 0/1, and the rest (0/3-0/24) should be clients for the second network that would be managed by the 2600 router. I configured everything with two VLANs: 100 for outside (0/2-0/24), 200 for inside (0/1-0/2). 0/2 is a trunk port for the two VLANs. The issue that came about is that I can't have two VLANs on at once: software doesn't allow it. Now, I can ping the outside network devices (172.16.7.1, 172.16.7.103), and even google (8.8.8.8) from the router, but not the switch. Devices on connected get a DHCP lease properly but can't ping outside the network, just the router - 172.17.7.1 and the switch itself, 172.17.7.7. The configuration for both the router and the switch are here, as well as below. Router: rt.throom#sho run Building configuration... Current configuration : 1015 bytes ! version 12.1 no service single-slot-reload-enable service timestamps debug uptime service timestamps log uptime no service password-encryption ! hostname rt.throom ! enable password To053cret ! ! ! ! ! no ip subnet-zero ip dhcp excluded-address 172.17.7.1 172.17.7.2 ip dhcp excluded-address 172.17.7.3 172.17.7.4 ip dhcp excluded-address 172.17.7.5 ! ip dhcp pool VLAN200 network 172.17.7.0 255.255.255.0 default-router 172.17.7.1 dns-server 8.8.8.8 ! ip audit notify log ip audit po max-events 100 ! ! ! ! ! ! ! interface Ethernet0/0 no ip address ! interface Ethernet0/0.100 encapsulation dot1Q 100 ip address 172.16.7.15 255.255.255.0 ip nat outside ! interface Ethernet0/0.200 encapsulation dot1Q 200 ip address 172.17.7.1 255.255.255.0 ip nat inside ! router eigrp 20 network 172.16.0.0 network 172.17.0.0 no auto-summary no eigrp log-neighbor-changes ! no ip classless no ip http server ! access-list 1 permit 172.17.7.0 0.0.0.255 ! ! line con 0 line aux 0 line vty 0 4 login ! end Switch: sw.throom#sho run Building configuration... Current configuration: ! version 11.2 no service pad no service udp-small-servers no service tcp-small-servers ! hostname sw.throom ! enable password Oh5053cret ! ! no spanning-tree vlan 100 no spanning-tree vlan 200 ip subnet-zero ! ! interface VLAN1 no ip address no ip route-cache ! interface FastEthernet0/1 switchport access vlan 100 spanning-tree portfast ! interface FastEthernet0/2 switchport trunk encapsulation dot1q switchport mode trunk ! interface FastEthernet0/3 switchport access vlan 200 spanning-tree portfast ! interface FastEthernet0/4 switchport access vlan 200 spanning-tree portfast ! interface FastEthernet0/5 switchport access vlan 200 spanning-tree portfast ! interface FastEthernet0/6 switchport access vlan 200 spanning-tree portfast ! interface FastEthernet0/7 switchport access vlan 200 spanning-tree portfast ! interface FastEthernet0/8 switchport access vlan 200 spanning-tree portfast ! interface FastEthernet0/9 switchport access vlan 200 spanning-tree portfast ! interface FastEthernet0/10 switchport access vlan 200 spanning-tree portfast ! interface FastEthernet0/11 switchport access vlan 200 spanning-tree portfast ! interface FastEthernet0/12 switchport access vlan 200 spanning-tree portfast ! interface FastEthernet0/13 switchport access vlan 200 spanning-tree portfast ! interface FastEthernet0/14 switchport access vlan 200 spanning-tree portfast ! interface FastEthernet0/15 switchport access vlan 200 spanning-tree portfast ! interface FastEthernet0/16 switchport access vlan 200 spanning-tree portfast ! interface FastEthernet0/17 switchport access vlan 200 spanning-tree portfast ! interface FastEthernet0/18 switchport access vlan 200 spanning-tree portfast ! interface FastEthernet0/19 switchport access vlan 200 spanning-tree portfast ! interface FastEthernet0/20 switchport access vlan 200 spanning-tree portfast ! interface FastEthernet0/21 switchport access vlan 200 spanning-tree portfast ! interface FastEthernet0/22 switchport access vlan 200 spanning-tree portfast ! interface FastEthernet0/23 switchport access vlan 200 spanning-tree portfast ! interface FastEthernet0/24 switchport access vlan 200 spanning-tree portfast ! ! line con 0 stopbits 1 line vty 0 4 login line vty 5 9 login ! end sho ip route gives: Gateway of last resort is 172.16.7.1 to network 0.0.0.0 172.17.0.0/24 is subnetted, 1 subnets C 172.17.7.0 is directly connected, Ethernet0/0.200 172.16.0.0/24 is subnetted, 1 subnets C 172.16.7.0 is directly connected, Ethernet0/0.100 S* 0.0.0.0/0 [1/0] via 172.16.7.1

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  • Is this the right strategy to convert an in-level order binary tree to a doubly linked list?

    - by Ankit Soni
    So I recently came across this question - Make a function that converts a in-level-order binary tree into a doubly linked list. Apparently, it's a common interview question. This is the strategy I had - Simply do a pre-order traversal of the tree, and instead of returning a node, return a list of nodes, in the order in which you traverse them. i.e return a list, and append the current node to the list at each point. For the base case, return the node itself when you are at a leaf. so you would say left = recursive_function(node.left) right = recursive_function(node.right) return(left.append(node.data)).append(right);` Is this the right approach?

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  • adding nodes to a binary search tree randomly deletes nodes

    - by SDLFunTimes
    Hi, stack. I've got a binary tree of type TYPE (TYPE is a typedef of data*) that can add and remove elements. However for some reason certain values added will overwrite previous elements. Here's my code with examples of it inserting without overwriting elements and it not overwriting elements. the data I'm storing: struct data { int number; char *name; }; typedef struct data data; # ifndef TYPE # define TYPE data* # define TYPE_SIZE sizeof(data*) # endif The tree struct: struct Node { TYPE val; struct Node *left; struct Node *rght; }; struct BSTree { struct Node *root; int cnt; }; The comparator for the data. int compare(TYPE left, TYPE right) { int left_len; int right_len; int shortest_string; /* find longest string */ left_len = strlen(left->name); right_len = strlen(right->name); if(right_len < left_len) { shortest_string = right_len; } else { shortest_string = left_len; } /* compare strings */ if(strncmp(left->name, right->name, shortest_string) > 1) { return 1; } else if(strncmp(left->name, right->name, shortest_string) < 1) { return -1; } else { /* strings are equal */ if(left->number > right->number) { return 1; } else if(left->number < right->number) { return -1; } else { return 0; } } } And the add method struct Node* _addNode(struct Node* cur, TYPE val) { if(cur == NULL) { /* no root has been made */ cur = _createNode(val); return cur; } else { int cmp; cmp = compare(cur->val, val); if(cmp == -1) { /* go left */ if(cur->left == NULL) { printf("adding on left node val %d\n", cur->val->number); cur->left = _createNode(val); } else { return _addNode(cur->left, val); } } else if(cmp >= 0) { /* go right */ if(cur->rght == NULL) { printf("adding on right node val %d\n", cur->val->number); cur->rght = _createNode(val); } else { return _addNode(cur->rght, val); } } return cur; } } void addBSTree(struct BSTree *tree, TYPE val) { tree->root = _addNode(tree->root, val); tree->cnt++; } The function to print the tree: void printTree(struct Node *cur) { if (cur == 0) { printf("\n"); } else { printf("("); printTree(cur->left); printf(" %s, %d ", cur->val->name, cur->val->number); printTree(cur->rght); printf(")\n"); } } Here's an example of some data that will overwrite previous elements: struct BSTree myTree; struct data myData1, myData2, myData3; myData1.number = 5; myData1.name = "rooty"; myData2.number = 1; myData2.name = "lefty"; myData3.number = 10; myData3.name = "righty"; initBSTree(&myTree); addBSTree(&myTree, &myData1); addBSTree(&myTree, &myData2); addBSTree(&myTree, &myData3); printTree(myTree.root); Which will print: (( righty, 10 ) lefty, 1 ) Finally here's some test data that will go in the exact same spot as the previous data, but this time no data is overwritten: struct BSTree myTree; struct data myData1, myData2, myData3; myData1.number = 5; myData1.name = "i"; myData2.number = 5; myData2.name = "h"; myData3.number = 5; myData3.name = "j"; initBSTree(&myTree); addBSTree(&myTree, &myData1); addBSTree(&myTree, &myData2); addBSTree(&myTree, &myData3); printTree(myTree.root); Which prints: (( j, 5 ) i, 5 ( h, 5 ) ) Does anyone know what might be going wrong? Sorry if this post was kind of long.

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  • Tree Surgeon 2.0 - The future on the T4 Express

    - by Malcolm Anderson
    If you've never been a fan of TreeSurgeon (http://treesurgeon.codeplex.com/) then skip this post.However, if have been there have been some interesting developments over the last couple of years.The biggest one is T4Recently Bill Simser wrote a detailed post about the potential future of tree surgeon, called "Tree Surgeon - Alive and Kicking or Dead and Buried" He raised the question:Times have changed. Since that last release in 2008 so much has changed for .NET developers. The question is, today is the project still viable? Do we still need a tool to generate a project tree given that we have things like scaffolding systems, NuGet, and T4 templates. Or should we just give the project its rightful and respectful send off as its had a good life and has outlived its usefulness.For myself, the answer is, keep it.I've spent the last couple of years doing agile engineering coaching and architecture and from my experience, I can tell you, there are a lot of shops out there that would benefit from having Tree Surgeon as a viable product.  Many would benefit simply from having the software engineering information that is embedded in the tree surgeon site be floating around their conversation.Little things like, keep all of your software needed to run the build, with the build in the version control system.Have your developers and the build system using the same build.Have a one-touch buildSeparate your code from your interfacePut unit tests in first, not lastI've seen companies with great developers suffer from the problems that naturally come from builds taking 3 and 4 hours to run.  It takes work to get that build down to 10 minutes, but the benefits are always worth it.  Tree Surgeon gives you a leg up, by starting you off with a project that you can drop into your Continuous Integration system, right out of the box.Well, it used to be right out of the box.  Today, you have to play with the project to make it work for you, but even with the issues (it hasn't been updated since 2008) it still gives you a framework, with logical separations that you can build from.If you have used Tree Surgeon in the past, take a few minutes and drop a comment about what difference it made in your development style, and what you are doing differently today because of it.

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  • Worst Case number of rotations for BST to AVL algorithm?

    - by spacker_lechuck
    I have a basic algorithm below and I know that the worst case input BST is one that has degenerated to a linked list from inserts to only one side. How would I compute the worst case complexity in terms of number of rotations for this BST to AVL conversion algorithm? IF tree is right heavy { IF tree's right subtree is left heavy { Perform Double Left rotation } ELSE { Perform Single Left rotation } } ELSE IF tree is left heavy { IF tree's left subtree is right heavy { Perform Double Right rotation } ELSE { Perform Single Right rotation } }

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  • GWT: Change padding of tree rows?

    - by Epaga
    A GWT tree looks roughly like this: <div class="gwt-Tree"> <div style="padding-top: 3px; padding-right: 3px; padding-bottom: 3px; margin-left: 0px; padding-left: 23px;"> <div style="display:inline;" class="gwt-TreeItem"> <table> ... </table> </div> </div> <div ...> </div> ... </div> My question is: how should I change the padding of the individual tree rows? I suppose I could do something along the lines of setting CSS rules for .gwt-Tree > div but that seems hacky. Is there a more elegant way?

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  • red black tree balancing?

    - by Anirudh Kaki
    i am working to generate tango tree, where i need to check whether every sub tree in tango is balanced or not. if its not balanced i need to make it balance? I trying so hard to make entire RB-tree balance but i not getting any proper logic so can any one help me out?? here i am adding code to check how to find my tree is balanced are not but when its not balanced how can i make it balance. static boolean verifyProperty5(rbnode n) { int left = 0, right = 0; if (n != null) { bh++; left = blackHeight(n.left, 0); right = blackHeight(n.right, 0); } if (left == right) { System.out.println("black height is :: " + bh); return true; } else { System.out.println("in balance"); return false; } } public static int blackHeight(rbnode root, int len) { bh = 0; blackHeight(root, path1, len); return bh; } private static void blackHeight(rbnode root, int path1[], int len) { if (root == null) return; if (root.color == "black"){ root.black_count = root.parent.black_count+1; } else{ root.black_count = root.parent.black_count; } if ((root.left == null) && (root.right == null)) { bh = root.black_count; } blackHeight(root.left, path1, len); blackHeight(root.right, path1, len); }

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