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  • transform file/directory structure into 'tree' in javascript

    - by dave
    I have an array of objects that looks like this: [{ name: 'test', size: 0, type: 'directory', path: '/storage/test' }, { name: 'asdf', size: 170, type: 'directory', path: '/storage/test/asdf' }, { name: '2.txt', size: 0, type: 'file', path: '/storage/test/asdf/2.txt' }] There could be any number of arbitrary path's, this is the result of iterating through files and folders within a directory. What I'm trying to do is determine the 'root' node of these. Ultimately, this will be stored in mongodb and use materialized path to determine it's relationships. In this example, /storage/test is a root with no parent. /storage/test/asdf has the parent of /storage/test which is the parent to /storage/test/asdf/2.txt. My question is, how would you go about iterating through this array, to determine the parent's and associated children? Any help in the right direction would be great! Thank you

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  • Javascript / Jquery Tree Travesal question

    - by Copper
    Suppose I have the following <ul> <li>Item 1</li> <li>Item 2 <ul> <li>Sub Item</li> </ul> </li> <li>Item 3</li> </ul> This list is auto-generated by some other code (so adding exclusive id's/class' is out of the question. Suppose I have some jquery code that states that if I mouseover an li, it gets a background color. However, if I mouseover the "Sub Item" list item, "Item 2" will be highlighted as well. How can I make it so that if the user mouses over "Sub Item" it only puts a background color on that and not on "Item 2" as well?

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  • How to find Sub-trees in non-binary tree

    - by kenny
    I have a non-binary tree. I want to find all "sub-trees" that are connected to root. Sub-tree is a a link group of tree nodes. every group is colored in it's own color. What would be be the best approach? Run recursion down and up for every node? The data structure of every treenode is a list of children, list of parents. (the type of children and parents are treenodes) Clarification: Group defined if there is a kind of "closure" between nodes where root itself is not part of the closure. As you can see from the graph you can't travel from pink to other nodes (you CAN NOT use root). From brown node you can travel to it's child so this form another group. Finally you can travel from any cyan node to other cyan nodes so the form another group

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  • GUI question : representing large tree

    - by Peter
    I have a tree-like datastructure of some six levels deep, that I would like to represent on a single webpage (can be tabs, trees; ....) In each level both childnodes and content are possible. Presenting it like a real tree would be not very usable (too big). I was thinking in the lines of hiding parts of the tree when you drill down and presenting a breadcrumbs or the like to keep you informed as to where you are... I guess my question boils down to : any ideas / examples ? Tx!

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  • How to find siblings of a tree?

    - by smallB
    On my interview for an internship, I was asked following question: On a whiteboard write the simplest algorithm with use of recursion which would take a root of a so called binary tree (so called because it is not strictly speaking binary tree) and make every child in this tree connected with its sibling. So if I have: 1 / \ 2 3 / \ \ 4 5 6 / \ 7 8 then the sibling to 2 would be 3, to four five, to five six and to seven eight. I didn't do this, although I was heading in the right direction. Later (next day) at home I did it, but with the use of a debugger. It took me better part of two hours and 50 lines of code. I personally think that this was very difficult question, almost impossible to do correctly on a whiteboard. How would you solve it on a whiteboard? How to apprehend this question without using a debugger?

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  • Quick Outline: Navigating Your PL/SQL Packages in Oracle SQL Developer

    - by thatjeffsmith
    If you’re browsing your packages using the Connections panel, you have a nice tree navigator to click around your packages and your variable, procedure, and functions. Click, click, click all day long, click, click, click while I sing this song… But What if you drill into your PL/SQL source from the worksheet and don’t have the Tree expanded? Let’s say you’re working on your script, something like - Hmm, what goes next again? So I need to reacquaint myself with just what my beer package requires, so I’m going to drill into it by doing a DESCRIBE (via SHIFT+F4), and now I have the package open. The package is open but the tree hasn’t auto-expanded. Please don’t tell me I have to do the click-click-click thing in the tree!?! Just Open the Quick Outline Panel Do you see it? Just right click in the procedure editor – select the ‘Quick Outline’ in the context menu, and voila! The navigational power of the tree, without needing to drill down the tree itself. If I want to drill into my procedure declaration, just click on said procedure name in the Quick Outline panel. This works for both package specs and bodies. Technically you can use this for stand alone procedures and functions, but the real power is demonstrated for packages.

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  • Asymptotic runtime of list-to-tree function

    - by Deestan
    I have a merge function which takes time O(log n) to combine two trees into one, and a listToTree function which converts an initial list of elements to singleton trees and repeatedly calls merge on each successive pair of trees until only one tree remains. Function signatures and relevant implementations are as follows: merge :: Tree a -> Tree a -> Tree a --// O(log n) where n is size of input trees singleton :: a -> Tree a --// O(1) empty :: Tree a --// O(1) listToTree :: [a] -> Tree a --// Supposedly O(n) listToTree = listToTreeR . (map singleton) listToTreeR :: [Tree a] -> Tree a listToTreeR [] = empty listToTreeR (x:[]) = x listToTreeR xs = listToTreeR (mergePairs xs) mergePairs :: [Tree a] -> [Tree a] mergePairs [] = [] mergePairs (x:[]) = [x] mergePairs (x:y:xs) = merge x y : mergePairs xs This is a slightly simplified version of exercise 3.3 in Purely Functional Data Structures by Chris Okasaki. According to the exercise, I shall now show that listToTree takes O(n) time. Which I can't. :-( There are trivially ceil(log n) recursive calls to listToTreeR, meaning ceil(log n) calls to mergePairs. The running time of mergePairs is dependent on the length of the list, and the sizes of the trees. The length of the list is 2^h-1, and the sizes of the trees are log(n/(2^h)), where h=log n is the first recursive step, and h=1 is the last recursive step. Each call to mergePairs thus takes time (2^h-1) * log(n/(2^h)) I'm having trouble taking this analysis any further. Can anyone give me a hint in the right direction?

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  • Binary Search Tree, cannot do traversal

    - by ihm
    Please see BST codes below. It only outputs "5". what did I do wrong? #include <iostream> class bst { public: bst(const int& numb) : root(new node(numb)) {} void insert(const int& numb) { root->insert(new node(numb), root); } void inorder() { root->inorder(root); } private: class node { public: node(const int& numb) : left(NULL), right(NULL) { value = numb; } void insert(node* insertion, node* position) { if (position == NULL) position = insertion; else if (insertion->value > position->value) insert(insertion, position->right); else if (insertion->value < position->value) insert(insertion, position->left); } void inorder(node* tree) { if (tree == NULL) return; inorder(tree->left); std::cout << tree->value << std::endl; inorder(tree->right); } private: node* left; node* right; int value; }; node* root; }; int main() { bst tree(5); tree.insert(4); tree.insert(2); tree.insert(10); tree.insert(14); tree.inorder(); return 0; }

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  • How to cleanly add after-the-fact commits from the same feature into git tree

    - by Dennis
    I am one of two developers on a system. I make most of the commits at this time period. My current git workflow is as such: there is master branch only (no develop/release) I make a new branch when I want to do a feature, do lots of commits, and then when I'm done, I merge that branch back into master, and usually push it to remote. ...except, I am usually not done. I often come back to alter one thing or another and every time I think it is done, but it can be 3-4 commits before I am really done and move onto something else. Problem The problem I have now is that .. my feature branch tree is merged and pushed into master and remote master, and then I realize that I am not really done with that feature, as in I have finishing touches I want to add, where finishing touches may be cosmetic only, or may be significant, but they still belong to that one feature I just worked on. What I do now Currently, when I have extra after-the-fact commits like this, I solve this problem by rolling back my merge, and re-merging my feature branch into master with my new commits, and I do that so that git tree looks clean. One clean feature branch branched out of master and merged back into it. I then push --force my changes to origin, since my origin doesn't see much traffic at the moment, so I can almost count that things will be safe, or I can even talk to other dev if I have to coordinate. But I know it is not a good way to do this in general, as it rewrites what others may have already pulled, causing potential issues. And it did happen even with my dev, where git had to do an extra weird merge when our trees diverged. Other ways to solve this which I deem to be not so great Next best way is to just make those extra commits to the master branch directly, be it fast-forward merge, or not. It doesn't make the tree look as pretty as in my current way I'm solving this, but then it's not rewriting history. Yet another way is to wait. Maybe wait 24 hours and not push things to origin. That way I can rewrite things as I see fit. The con of this approach is time wasted waiting, when people may be waiting for a fix now. Yet another way is to make a "new" feature branch every time I realize I need to fix something extra. I may end up with things like feature-branch feature-branch-html-fix, feature-branch-checkbox-fix, and so on, kind of polluting the git tree somewhat. Is there a way to manage what I am trying to do without the drawbacks I described? I'm going for clean-looking history here, but maybe I need to drop this goal, if technically it is not a possibility.

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  • How can I implement a splay tree that performs the zig operation last, not first?

    - by Jakob
    For my Algorithms & Data Structures class, I've been tasked with implementing a splay tree in Haskell. My algorithm for the splay operation is as follows: If the node to be splayed is the root, the unaltered tree is returned. If the node to be splayed is one level from the root, a zig operation is performed and the resulting tree is returned. If the node to be splayed is two or more levels from the root, a zig-zig or zig-zag operation is performed on the result of splaying the subtree starting at that node, and the resulting tree is returned. This is valid according to my teacher. However, the Wikipedia description of a splay tree says the zig step "will be done only as the last step in a splay operation" whereas in my algorithm it is the first step in a splay operation. I want to implement a splay tree that performs the zig operation last instead of first, but I'm not sure how it would best be done. It seems to me that such an algorithm would become more complex, seeing as how one needs to find the node to be splayed before it can be determined whether a zig operation should be performed or not. How can I implement this in Haskell (or some other functional language)?

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  • Sentence Tree v/s Words List

    - by Rohit Jose
    I was recently tasked with building a Name Entity Recognizer as part of a project. The objective was to parse a given sentence and come up with all the possible combinations of the entities. One approach that was suggested was to keep a lookup table for all the know connector words like articles and conjunctions, remove them from the words list after splitting the sentence on the basis of the spaces. This would leave out the Name Entities in the sentence. A lookup is then done for these identified entities on another lookup table that associates them to the entity type, for example if the sentence was: Remember the Titans was a movie directed by Boaz Yakin, the possible outputs would be: {Remember the Titans,Movie} was {a movie,Movie} directed by {Boaz Yakin,director} {Remember the Titans,Movie} was a movie directed by Boaz Yakin {Remember the Titans,Movie} was {a movie,Movie} directed by Boaz Yakin {Remember the Titans,Movie} was a movie directed by {Boaz Yakin,director} Remember the Titans was {a movie,Movie} directed by Boaz Yakin Remember the Titans was {a movie,Movie} directed by {Boaz Yakin,director} Remember the Titans was a movie directed by {Boaz Yakin,director} Remember the {the titans,Movie,Sports Team} was {a movie,Movie} directed by {Boaz Yakin,director} Remember the {the titans,Movie,Sports Team} was a movie directed by Boaz Yakin Remember the {the titans,Movie,Sports Team} was {a movie,Movie} directed by Boaz Yakin Remember the {the titans,Movie,Sports Team} was a movie directed by {Boaz Yakin,director} The entity lookup table here would contain the following data: Remember the Titans=Movie a movie=Movie Boaz Yakin=director the Titans=Movie the Titans=Sports Team Another alternative logic that was put forward was to build a crude sentence tree that would contain the connector words in the lookup table as parent nodes and do a lookup in the entity table for the leaf node that might contain the entities. The tree that was built for the sentence above would be: The question I am faced with is the benefits of the two approaches, should I be going for the tree approach to represent the sentence parsing, since it provides a more semantic structure? Is there a better approach I should be going for solving it?

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  • vector rotations for branches of a 3d tree

    - by freefallr
    I'm attempting to create a 3d tree procedurally. I'm hoping that someone can check my vector rotation maths, as I'm a bit confused. I'm using an l-system (a recursive algorithm for generating branches). The trunk of the tree is the root node. It's orientation is aligned to the y axis. In the next iteration of the tree (e.g. the first branches), I might create a branch that is oriented say by +10 degrees in the X axis and a similar amount in the Z axis, relative to the trunk. I know that I should keep a rotation matrix at each branch, so that it can be applied to child branches, along with any modifications to the child branch. My questions then: for the trunk, the rotation matrix - is that just the identity matrix * initial orientation vector ? for the first branch (and subsequent branches) - I'll "inherit" the rotation matrix of the parent branch, and apply x and z rotations to that also. e.g. using glm::normalize; using glm::rotateX; using glm::vec4; using glm::mat4; using glm::rotate; vec4 vYAxis = vec4(0.0f, 1.0f, 0.0f, 0.0f); vec4 vInitial = normalize( rotateX( vYAxis, 10.0f ) ); mat4 mRotation = mat4(1.0); // trunk rotation matrix = identity * initial orientation vector mRotation *= vInitial; // first branch = parent rotation matrix * this branches rotations mRotation *= rotate( 10.0f, 1.0f, 0.0f, 0.0f ); // x rotation mRotation *= rotate( 10.0f, 0.0f, 0.0f, 1.0f ); // z rotation Are my maths and approach correct, or am I completely wrong? Finally, I'm using the glm library with OpenGL / C++ for this. Is the order of x rotation and z rotation important?

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  • Cloning from a given point in the snapshot tree

    - by Fat Bloke
    Although we have just released VirtualBox 4.3, this quick blog entry is about a longer standing ability of VirtualBox when it comes to Snapshots and Cloning, and was prompted by a question posed internally, here in Oracle: "Is there a way I can create a new VM from a point in my snapshot tree?". Here's the scenario: Let's say you have your favourite work VM which is Oracle Linux based and as you installed different packages, such as database, middleware, and the apps, you took snapshots at each point like this: But you then need to create a new VM for some other testing or to share with a colleague who will be using the same Linux and Database layers but may want to reconfigure the Middleware tier, and may want to install his own Apps. All you have to do is right click on the snapshot that you're happy with and clone: Give the VM that you are about to create a name, and if you plan to use it on the same host machine as the original VM, it's a good idea to "Reinitialize the MAC address" so there's no clash on the same network: Now choose the Clone type. If you plan to use this new VM on the same host as the original, you can use Linked Cloning else choose Full.  At this point you now have a choice about what to do about your snapshot tree. In our example, we're happy with the Linux and Database layers, but we may want to allow our colleague to change the upper tiers, with the option of reverting back to our known-good state, so we'll retain the snapshot data in the new VM from this point on: The cloning process then chugs along and may take a while if you chose a Full Clone: Finally, the newly cloned VM is ready with the subset of the Snapshot tree that we wanted to retain: Pretty powerful, and very useful.  Cheers, -FB 

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  • Return the difference between the lowest and highest key

    - by stan
    This is a past exam paper i am attempting and have no way to check if the out put is correct as i am not capable of building one of these things the question is in the title class Tree{ Tree left; Tree right; int key; public static int span(Tree tree) { if ( tree == null ){ return null; } if( tree.left != null) int min = span(tree.left); } if( tree.right != null){ int max = span(tree.right); } return max - min; } } Could anyone suggest what i need to change to get 5/5 marks :D - the only thing we have to do is write the span method, the header was given for us Thanks

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  • Visual Tree Enumeration

    - by codingbloke
    I feel compelled to post this blog because I find I’m repeatedly posting this same code in silverlight and windows-phone-7 answers in Stackoverflow. One common task that we feel we need to do is burrow into the visual tree in a Silverlight or Windows Phone 7 application (actually more recently I found myself doing this in WPF as well).  This allows access to details that aren’t exposed directly by some controls.  A good example of this sort of requirement is found in the “Restoring exact scroll position of a listbox in Windows Phone 7”  question on stackoverflow.  This required that the scroll position of the scroll viewer internal to a listbox be accessed. A caveat One caveat here is that we should seriously challenge the need for this burrowing since it may indicate that there is a design problem.  Burrowing into the visual tree or indeed burrowing out to containing ancestors could represent significant coupling between module boundaries and that generally isn’t a good idea. Why isn’t this idea just not cast aside as a no-no?  Well the whole concept of a “Templated Control”, which are in extensive use in these applications, opens the coupling between the content of the visual tree and the internal code of a control.   For example, I can completely change the appearance and positioning of elements that make up a ComboBox.  The ComboBox control relies on specific template parts having set names of a specified type being present in my template.  Rightly or wrongly this does kind of give license to writing code that has similar coupling. Hasn’t this been done already? Yes it has.  There are number of blogs already out there with similar solutions.  In fact if you are using Silverlight toolkit the VisualTreeExtensions class already provides this feature.  However I prefer my specific code because of the simplicity principle I hold to.  Only write the minimum code necessary to give all the features needed.  In this case I add just two extension methods Ancestors and Descendents, note I don’t bother with “Get” or “Visual” prefixes.  Also I haven’t added Parent or Children methods nor additional “AndSelf” methods because all but Children is achievable with the addition of some other Linq methods.  I decided to give Descendents an additional overload for depth hence a depth of 1 is equivalent to Children but this overload is a little more flexible than simply Children. So here is the code:- VisualTreeEnumeration public static class VisualTreeEnumeration {     public static IEnumerable<DependencyObject> Descendents(this DependencyObject root, int depth)     {         int count = VisualTreeHelper.GetChildrenCount(root);         for (int i = 0; i < count; i++)         {             var child = VisualTreeHelper.GetChild(root, i);             yield return child;             if (depth > 0)             {                 foreach (var descendent in Descendents(child, --depth))                     yield return descendent;             }         }     }     public static IEnumerable<DependencyObject> Descendents(this DependencyObject root)     {         return Descendents(root, Int32.MaxValue);     }     public static IEnumerable<DependencyObject> Ancestors(this DependencyObject root)     {         DependencyObject current = VisualTreeHelper.GetParent(root);         while (current != null)         {             yield return current;             current = VisualTreeHelper.GetParent(current);         }     } }   Usage examples The following are some examples of how to combine the above extension methods with Linq to generate the other axis scenarios that tree traversal code might require. Missing Axis Scenarios var parent = control.Ancestors().Take(1).FirstOrDefault(); var children = control.Descendents(1); var previousSiblings = control.Ancestors().Take(1)     .SelectMany(p => p.Descendents(1).TakeWhile(c => c != control)); var followingSiblings = control.Ancestors().Take(1)     .SelectMany(p => p.Descendents(1).SkipWhile(c => c != control).Skip(1)); var ancestorsAndSelf = Enumerable.Repeat((DependencyObject)control, 1)     .Concat(control.Ancestors()); var descendentsAndSelf = Enumerable.Repeat((DependencyObject)control, 1)     .Concat(control.Descendents()); You might ask why I don’t just include these in the VisualTreeEnumerator.  I don’t on the principle of only including code that is actually needed.  If you find that one or more of the above  is needed in your code then go ahead and create additional methods.  One of the downsides to Extension methods is that they can make finding the method you actually want in intellisense harder. Here are some real world usage scenarios for these methods:- Real World Scenarios //Gets the internal scrollviewer of a ListBox ScrollViewer sv = someListBox.Descendents().OfType<ScrollViewer>().FirstOrDefault(); // Get all text boxes in current UserControl:- var textBoxes = this.Descendents().OfType<TextBox>(); // All UIElement direct children of the layout root grid:- var topLevelElements = LayoutRoot.Descendents(0).OfType<UIElement>(); // Find the containing `ListBoxItem` for a UIElement:- var container = elem.Ancestors().OfType<ListBoxItem>().FirstOrDefault(); // Seek a button with the name "PinkElephants" even if outside of the current Namescope:- var pinkElephantsButton = this.Descendents()     .OfType<Button>()     .FirstOrDefault(b => b.Name == "PinkElephants"); //Clear all checkboxes with the name "Selector" in a Treeview foreach (CheckBox checkBox in elem.Descendents()     .OfType<CheckBox>().Where(c => c.Name == "Selector")) {     checkBox.IsChecked = false; }   The last couple of examples above demonstrate a common requirement of finding controls that have a specific name.  FindName will often not find these controls because they exist in a different namescope. Hope you find this useful, if not I’m just glad to be able to link to this blog in future stackoverflow answers.

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  • How do I optimize searching for the nearest point?

    - by Rootosaurus
    For a little project of mine I'm trying to implement a space colonization algorithm in order to grow trees. The current implementation of this algorithm works fine. But I have to optimize the whole thing in order to make it generate faster. I work with 1 to 300K of random attraction points to generate one tree, and it takes a lot of time to compute and compare distances between attraction points and tree node in order to keep only the closest treenode for an attraction point. So I was wondering if some solutions exist (I know they must exist) in order to avoid the time loss looping on each tree node for each attraction point to find the closest... and so on until the tree is finished.

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  • reconstructing a tree from its preorder and postorder lists.

    - by NomeN
    Consider the situation where you have two lists of nodes of which all you know is that one is a representation of a preorder traversal of some tree and the other a representation of a postorder traversal of the same tree. I believe it is possible to reconstruct the tree exactly from these two lists, and I think I have an algorithm to do it, but have not proven it. As this will be a part of a masters project I need to be absolutely certain that it is possible and correct (Mathematically proven). However it will not be the focus of the project, so I was wondering if there is a source out there (i.e. paper or book) I could quote for the proof. (Maybe in TAOCP? anybody know the section possibly?) In short, I need a proven algorithm in a quotable resource that reconstructs a tree from its pre and post order traversals. Note: The tree in question will probably not be binary, or balanced, or anything that would make it too easy. Note2: Using only the preorder or the postorder list would be even better, but I do not think it is possible. Note3: A node can have any amount of children. Note4: I only care about the order of siblings. Left or right does not matter when there is only one child.

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  • What is the most efficient/elegant way to parse a flat table into a tree?

    - by Tomalak
    Assume you have a flat table that stores an ordered tree hierarchy: Id Name ParentId Order 1 'Node 1' 0 10 2 'Node 1.1' 1 10 3 'Node 2' 0 20 4 'Node 1.1.1' 2 10 5 'Node 2.1' 3 10 6 'Node 1.2' 1 20 What minimalistic approach would you use to output that to HTML (or text, for that matter) as a correctly ordered, correctly intended tree? Assume further you only have basic data structures (arrays and hashmaps), no fancy objects with parent/children references, no ORM, no framework, just your two hands. The table is represented as a result set, which can be accessed randomly. Pseudo code or plain English is okay, this is purely a conceptional question. Bonus question: Is there a fundamentally better way to store a tree structure like this in a RDBMS? EDITS AND ADDITIONS To answer one commenter's (Mark Bessey's) question: A root node is not necessary, because it is never going to be displayed anyway. ParentId = 0 is the convention to express "these are top level". The Order column defines how nodes with the same parent are going to be sorted. The "result set" I spoke of can be pictured as an array of hashmaps (to stay in that terminology). For my example was meant to be already there. Some answers go the extra mile and construct it first, but thats okay. The tree can be arbitrarily deep. Each node can have N children. I did not exactly have a "millions of entries" tree in mind, though. Don't mistake my choice of node naming ('Node 1.1.1') for something to rely on. The nodes could equally well be called 'Frank' or 'Bob', no naming structure is implied, this was merely to make it readable. I have posted my own solution so you guys can pull it to pieces.

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  • What data is actually stored in a B-tree database in CouchDB?

    - by Andrey Vlasovskikh
    I'm wondering what is actually stored in a CouchDB database B-tree? The CouchDB: The Definitive Guide tells that a database B-tree is used for append-only operations and that a database is stored in a single B-tree (besides per-view B-trees). So I guess the data items that are appended to the database file are revisions of documents, not the whole documents: +---------|### ... | | +------|###|------+ ... ---+ | | | | +------+ +------+ +------+ +------+ | doc1 | | doc2 | | doc1 | ... | doc1 | | rev1 | | rev1 | | rev2 | | rev7 | +------+ +------+ +------+ +------+ Is it true? If it is true, then how the current revision of a document is determined based on such a B-tree? Doesn't it mean, that CouchDB needs a separate "view" database for indexing current revisions of documents to preserve O(log n) access? Wouldn't it lead to race conditions while building such an index? (as far as I know, CouchDB uses no write locks).

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  • Help me understand Inorder Traversal without using recursion

    - by vito
    I am able to understand preorder traversal without using recursion, but I'm having a hard time with inorder traversal. I just don't seem to get it, perhaps, because I haven't understood the inner working of recursion. This is what I've tried so far: def traverseInorder(node): lifo = Lifo() lifo.push(node) while True: if node is None: break if node.left is not None: lifo.push(node.left) node = node.left continue prev = node while True: if node is None: break print node.value prev = node node = lifo.pop() node = prev if node.right is not None: lifo.push(node.right) node = node.right else: break The inner while-loop just doesn't feel right. Also, some of the elements are getting printed twice; may be I can solve this by checking if that node has been printed before, but that requires another variable, which, again, doesn't feel right. Where am I going wrong? I haven't tried postorder traversal, but I guess it's similar and I will face the same conceptual blockage there, too. Thanks for your time! P.S.: Definitions of Lifo and Node: class Node: def __init__(self, value, left=None, right=None): self.value = value self.left = left self.right = right class Lifo: def __init__(self): self.lifo = () def push(self, data): self.lifo = (data, self.lifo) def pop(self): if len(self.lifo) == 0: return None ret, self.lifo = self.lifo return ret

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