Search Results

Search found 4232 results on 170 pages for 'tree traversal'.

Page 5/170 | < Previous Page | 1 2 3 4 5 6 7 8 9 10 11 12  | Next Page >

• 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 :).

  Read the article

• 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?

  Read the article

• 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);

  Read the article

• Why does this B+ tree have repeated elements?

  - by Phenom

  In this B+ tree 5 appears twice.

  Read the article

• 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)?

  Read the article

• 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?

  Read the article

• CVE-2012-1712 Path Traversal Vulnerability in Sun GlassFish Web Space Server

  - by chandan

  CVE DescriptionCVSSv2 Base ScoreComponentProduct and Resolution CVE-2012-1712 Path Traversal Vulnerability 5.0 Liferay Sun GlassFish Web Space Server 10.0 Update 7 Patch 2 SPARC X86 Linux Windows Oracle acknowledges with thanks, Christian Schneider for bringing this issue to our attention. This notification describes vulnerabilities fixed in third-party components that are included in Sun's product distribution.Information about vulnerabilities affecting Oracle Sun products can be found on Oracle Critical Patch Updates and Security Alerts page.

  Read the article

• CVE-2011-2524 Directory traversal vulnerability in libsoup

  - by RitwikGhoshal

  CVE DescriptionCVSSv2 Base ScoreComponentProduct and Resolution CVE-2011-2524 Improper Limitation of a Pathname to a Restricted Directory ('Path Traversal') vulnerability 5.0 libsoup Solaris 11 11/11 SRU 11.4 This notification describes vulnerabilities fixed in third-party components that are included in Oracle's product distributions.Information about vulnerabilities affecting Oracle products can be found on Oracle Critical Patch Updates and Security Alerts page.

  Read the article

• 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.

  Read the article

• 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.

  Read the article

• 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 } }

  Read the article

• 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?

  Read the article

• 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); }

  Read the article

• C++ find largest BST in a binary tree

  - by fonjibe

  what is your approach to have the largest BST in a binary tree? I refer to this post where a very good implementation for finding if a tree is BST or not is bool isBinarySearchTree(BinaryTree * n, int min=std::numeric_limits<int>::min(), int max=std::numeric_limits<int>::max()) { return !n || (min < n->value && n->value < max && isBinarySearchTree(n->l, min, n->value) && isBinarySearchTree(n->r, n->value, max)); } It is quite easy to implement a solution to find whether a tree contains a binary search tree. i think that the following method makes it: bool includeSomeBST(BinaryTree* n) { if(!isBinarySearchTree(n)) { if(!isBinarySearchTree(n->left)) return isBinarySearchTree(n->right); } else return true; else return true; } but what if i want the largest BST? this is my first idea, BinaryTree largestBST(BinaryTree* n) { if(isBinarySearchTree(n)) return n; if(!isBinarySearchTree(n->left)) { if(!isBinarySearchTree(n->right)) if(includeSomeBST(n->right)) return largestBST(n->right); else if(includeSomeBST(n->left)) return largestBST(n->left); else return NULL; else return n->right; } else return n->left; } but its not telling the largest actually. i struggle to make the comparison. how should it take place? thanks

  Read the article

• Why can't RB-Tree be a list?

  - by Alex

  Hey everyone. I have a problem with the rb-trees. according to wikipedia, rb-tree needs to follow the following: A node is either red or black. The root is black. (This rule is used in some definitions and not others. Since the root can always be changed from red to black but not necessarily vice-versa this rule has little effect on analysis.) All leaves are black. Both children of every red node are black. Every simple path from a given node to any of its descendant leaves contains the same number of black nodes. As we know, an rb-tree needs to be balanced and has the height of O(log(n)). But, if we insert an increasing series of numbers (1,2,3,4,5...) and theoretically we will get a tree that will look like a list and will have the height of O(n) with all its nodes black, which doesn't contradict the rb-tree properties mentioned above. So, where am I wrong?? thanks.

  Read the article

• navigate all items in a wpf tree view

  - by Brian Leahy

  I want to be able to traverse the visual ui tree looking for an element with an ID bound to the visual element's Tag property. I'm wondering how i do this. Controls don't have children to traverse. I started using LogicalTreeHelper.GetChildren, which seems to work as intended, up until i hit a TreeView control... then LogicalTreeHelper.GetChildren doesnt return any children. Note: the purpose is to find the visual UI element that corresponds to the data item. That is, given an ID of the item, Go find the UI element displaying it. Edit: I am apparently am not explaining this well enough. I am binding some data objects to a TreeView control and then wanting to select a specific item programaticly given that business object's ID. I dont see why it's so hard to travers the visual tree and find the element i want, as the data object's ID is in the Tag property of the appropriate visual element. I'm using Mole and I am able to find the UI element with the appropriate ID in it's Tag. I just cannot find the visual element in code. LogicalTreeHelper does not traverse any items in the tree. Neither does ItemContainerGenerator.ContainerFromItem retrieve anything for items in the tree view.

  Read the article

• Evaluate an expression tree

  - by Phronima

  Hi, This project that I'm working on requires that an expression tree be constructed from a string of single digit operands and operators both represented as type char. I did the implmentation and the program up to that point works fine. I'm able to print out the inorder, preorder and postorder traversals in the correct way. The last part calls for evaulating the expression tree. The parameters are an expression tree "t" and its root "root". The expression tree is ((3+2)+(6+2)) which is equal to 13. Instead I get 11 as the answer. Clearly I'm missing something here and I've done everything short of bashing my head against the desk. I would greatly appreciate it if someone can point me in the right direction. (Note that at this point I'm only testing addition and will add in the other operators when I get this method working.) public int evalExpression( LinkedBinaryTree t, BTNode root ) { if( t.isInternal( root ) ) { int x = 0, y = 0, value = 0; char operator = root.element(); if( root.getLeft() != null ) x = evalExpression(t, t.left( root ) ); if( root.getRight() != null ) y = evalExpression(t, t.right( root ) ); if( operator == '+' ) { value = value + Character.getNumericValue(x) + Character.getNumericValue(y); } return value; } else { return root.element(); } }

  Read the article

• Flex Tree Properties, Null Reference?

  - by mvrak

  I am pulling down a large XML file and I have no control over it's structure. I used a custom function to use the tag name to view the tree structure as a flex tree, but then it breaks. I am guessing it has something to do with my other function, one that calls attribute values from the selected node. See code. <mx:Tree x="254" y="21" width="498" height="579" id="xmllisttree" labelFunction="namer" dataProvider="{treeData}" showRoot="false" change="treeChanged(event)" /> //and the Cdata import mx.rpc.events.ResultEvent; [Bindable] private var fullXML:XMLList; private function contentHandler(evt:ResultEvent):void{ fullXML = evt.result.page; } [Bindable] public var selectedNode:Object; public function treeChanged(event:Event):void { selectedNode=Tree(event.target).selectedItem; } public function namer(item:Object):String { var node:XML = XML(item); var nodeName:QName = node.name(); var stringtest:String ="bunny"; return nodeName.localName; } The error is TypeError: Error #1009: Cannot access a property or method of a null object reference. Where is the null reference?

  Read the article

• R Tree 50,000 foot overview?

  - by roufamatic

  I'm working on a school project that involves taking a lat/long point and finding the top five closest points in a known list of places. The list is to be stored in memory, with the caveat that we must choose an "appropriate data structure" -- that is, we cannot simply store all the places in an array and compare distances one-by-one in a linear fashion. The teacher suggested grouping the place data by US State to prevent calculating the distance for places that are obviously too far away. I think I can do better. From my research online it seems like an R-Tree or one of its variants might be a neat solution. Unfortunately, that sentence is as far as I've gotten with understanding the actual technique, as the literature is simply too dense for my non-academic head. Can somebody give me a really high overview of what the process is for populating an R-Tree with lat/long data, and then traversing the tree to find those 5 nearest neighbors of a given point? Additionally the project is in C, and I don't have to reinvent the wheel on this, so if you've used an existing open source C implementation of an R Tree I'd be interested in your experiences.

  Read the article

• Sorting by some 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 only 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). 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) After that for each result which is 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 thing?

  Read the article

• python recursive iteration exceeding limit for tree implementation

  - by user3698027

  I'm implementing a tree dynamically in python. I have defined a class like this... class nodeobject(): def __init__(self,presentnode=None,parent=None): self.currentNode = presentnode self.parentNode = parent self.childs = [] I have a function which gets possible childs for every node from a pool def findchildren(node, childs): # No need to write the whole function on how it gets childs Now I have a recursive function that starts with the head node (no parent) and moves down the chain recursively for every node (base case being the last node having no children) def tree(dad,children): for child in children: childobject = nodeobject(child,dad) dad.childs.append(childobject) newchilds = findchildren(child, children) if len(newchilds) == 0: lastchild = nodeobject(newchilds,childobject) childobject.childs.append(lastchild) loopchild = copy.deepcopy(lastchild) while loopchild.parentNode != None: print "last child" else: tree(childobject,newchilds) The tree formation works for certain number of inputs only. Once the pool gets bigger, it results into "MAXIMUM RECURSION DEPTH EXCEEDED" I have tried setting the recursion limit with set.recursionlimit() and it doesn't work. THe program crashes. I want to implement a stack for recursion, can someone please help, I have gone no where even after trying for a long time ?? Also, is there any other way to fix this other than stack ?

  Read the article

• Simulation tree command in C

  - by Ecle

  I have to create the simulation of tree command in C, this is my current code: #include <stdio.h> #include <sys/types.h> #include <sys/stat.h> #include <dirent.h> #include <string.h> main(int argc, char *argv[]){ int i; if(argc < 2){ printf("\nError. Use: %s directory\n", argv[0]); system("exit"); } for(i=1;i<argc;i++) //if(argv[i][0] != '-') tree(argv[i]); } tree(char *ruta){ DIR *dirp; struct dirent *dp; static nivel = 0; struct stat buf; char fichero[256]; int i; if((dirp = opendir(path)) == NULL){ perror(path); return; } while((dp = readdir(dirp)) != NULL){ printf(fichero, "%s/%s", path, dp->d_name); if((buf.st_mode & S_IFMT) == S_IFDIR){ for(i=0;i<nivel;i++) printf("\t"); printf("%s\n", dp->d_name); ++nivel; tree(fichero); --nivel; } } } Apparently, it works! (due to it compiles correctly) But I don't why. I can't pass the correct arguments to execute this. Thank you so much, people.

  Read the article

• Dynamic Tree control in Flex 3

  - by nimmyliji

  I am looking for sample code to create a dynamic Tree control in Flex using a Collection of objects obtained from the backend(Perl cgi). So, initially the Tree will display the root nodes. Clicking root node, will invoke the data for populating the child nodes (basically adding child nodes on demand). Clicking child nodes will pull another collection add child nodes of child node. So, lets assume Initially the Tree will display - Root1 Root2 Root3 Clicking Root1 will display something like this - Root1 Child 1 Child 2 Root2 Root3 And Clicking Child1 will display - Root1 Child 1 Child1 of Child 1 Child2 of Child 1 Child 2 Root2 Root3 Is it possible? Thanks in advance...

  Read the article

• Metalanguage like BNF or XML-Schema to validate a tree-instance against a tree-model

  - by Stefan

  Hi! I'm implementing a new machine learning algorithm in Java that extracts a prototype datastructure from a set of structured datasets (tree-structure). As im developing a generic library for that purpose, i kept my design independent from concrete data-representations like XML. My problem now is that I need a way to define a data model, which is basically a ruleset describing valid trees, against which a set of trees is being matched. I thought of using BNF or a similar dialect. Basically I need a way to iterate through the space of all valid TreeNodes defined by the ModelTree (Like a search through the search space for algorithms like A*) so that i can compare my set of concrete trees with the model. I know that I'll have to deal with infinite spaces there but first things first. I know, it's rather tricky (and my sentences are pretty bumpy) but I would appreciate any clues. Thanks in advance, Stefan

  Read the article

• how to handle duplicates in AVL tree

  - by Ahmed Kotb

  I want to make my avl tree support adding duplicates but there is a problem with the default behavior of the binary search tree with duplicates that the rotation could make nodes with equal key be on the left and the right of the parent for example adding A,A,A will cause the tree to do a rotation to be something like that A / \ A A so getting all the entries with that key will be a problem...and searching in both direction is not something good the solution that i have thought of is making each node stores an array of duplicates so when adding a new item that already exists will be just adding a new item to the array , removing item with key will delete the whole node while the find all items with key will return that array. are there are any other methods to handle duplicates ?

  Read the article

< Previous Page | 1 2 3 4 5 6 7 8 9 10 11 12  | Next Page >