There is a forest of doubly linked lists. Reshuffle it in place to get a  single doubly linked list?
534848
I remember coming across such a question one time, but I think it talked of converting a tree (and not a forest) into a linear list.\n\nI suggest you do the following: \n- Iterate through the forest, 1 tree at a time\n- for each tree, do an inorder traversal to generate a linear list (*)\n- join the linear lists to get the complete linear list\n\n(*) I guess this point needs more clarification\nWhat I am suggesting is this\n.      ROOT\n.       //   &#92;&#92;\n.   LEFT   RIGHT\nAssume this to be a trivial doubly linked tree\nAn inorder traversal can convert this into this structure\nLEFT=ROOT=RIGHT which is a doubly linked list really.\n\nI am not sure how to extend this to trees with more than 2 children, but I hope this helps!
606674
First of all add a treeview form to your form... \n\nfrom there I recommend using the file system object to get the hard drives from the system and all all of them as nodes, then when a folder is clicked, make it a child node of the parent folder...ect.\n\nTry a yahoo search for information on the File System object and tree nodes if you do not have any practical examples.....\n\nthanks
636496
Sure. You just need a set of vertices and a set of edges (directed edges in case of a directed graph). There's no requirement for the set of edges to be non-empty. \n\nSometimes even an empty set of vertices is allowed.
730795
Here are "drawings" or descriptions of all 11 trees:\n\n_ _ _ _ _ _\n\n_ _ _ _ L\n\n_ _ _ L _\n\n_ L _ L\n\n_ _ L L\n\nX _ _\n\n_ X _\n\nY  (where each of the three spokes consists of 2 edges)\n\n X.L\n\nA star with 5 spokes: one spoke consists of 2 edges:\n\n1-2 1-3 1-4 1-5 1-6 6-7\n\nA start with 6 spokes:\n\n1-2 1-3 1-4 1-5 1-6 1-7\n\nStart with the one tree with 0 edges. After drawing all the trees with n edges, add an extra edge to get all the trees with n+1 edges, removing any duplicates ("isomorphic" trees).
209333
no where.  Just the data pointers are deleted.
475021
Ok, basically your command is displaying a list of all files in the current directory based modification time that have a file name pattern of fork*.c where the * is any combination of characters followed by a period and then a c.    Next that listing is passed to the cut program which does something about further slimming down the results.  finally all the information is passed to sort which will return results removing any duplicates.    \n\nalright.. well with that said, hopefully that makes sense... now how you could simplify that into 2 commands is a bit beyond my know how but.. but i imagine you could probably use ls and grep only.  grep is a very powerful application and you can do many things with it.. i'd do some research on it and see what you can figure out.
517018
In order to understand the difference, you must also know what they have in common. Both of them are lists in which the length can vary. What I mean by that is that, unlike an array, which has a fixed length when it is first created, these lists can both "increase" in size when adding elements to them.\n\nThe difference is in how they implement it.\nAn ArrayList internally uses an array (or rather series of arrays) to store its elements. The ArrayList will make the determination whether to allocate more arrays in which to store elements when adding elements.\nA LinkedList stores its elements as a series of "node objects" of which each one has a reference to the element that comes after and the element that comes before (if any). (For the record, the Java implementation is a "Doubly linked list".)\nBoth have their advantages and disadvantages.\nArrayLists are very fast when it comes to finding elements based on index (because they use arrays internally). LinkedLists are much slower.\nLinkedLists are much more efficient at using memory (but in the Java implementation, this is not such a big deal). They are also more efficient at adding and removing elements from the beginning of the list, thus making them more suitable for stack implementations.
218696
There may be some good theory on this topic, but\nI'd like to point out one obvious solution and\none possibly-special-case solution.\n\nFirst, given a graph, if you have the ability to \ntraverse the entire graph and "mark" each node as\nyou visit it, then if you ever run into a marked\nnode, then there is a cycle.  The most obvious way\nto mark a node is to store a flag in it, but it\nis also possible to build a separate data structure\nto keep track of each node (ie. a hash table \nindexed by node address or something like that.)\n\nAnother trick I've seen used on lists, but *might* be\ngeneralized to some more-complicated graphs, is the\nidea of a follow-pointer.  In the case of a list,\nthe "follow pointer" traverses one list link\nfor every *two* that the real traversal covers.\nIf there is a loop in the list, eventually the\n"lead" pointer will wrap around and catch up\nto the "follow" pointer.  Thus, you know that\nthere's a cycle if the two pointers ever refer\nto the same node.
195503
There is no structure in Java.\n\nProbably you are confusing with C++>
438803
I've actually pondered this myself when taking digital circuit courses... The solution , i THINK, involves having not just one large memory unit , but instead, breaking the memory into 4 or 8 or 16 or more parts and using some kind of chip-select mechanism (using a multiplexer)
556841
Try Planet Source Code. Below are 52 examples (You probably should do your own homework, however...):\n\nhttp://www.planetsourcecode.com/vb/scripts/BrowseCategoryOrSearchResults.asp?blnWorldDropDownUsed=TRUE&txtMaxNumberOfEntriesPerPage=10&blnResetAllVariables=TRUE&optSort=Alphabetical&txtCriteria=Binary+Tree&lngWId=3
1116199
the only time you would need Trees is if your a programer then you would you use tree methods
687652
look at C++ book so many ... ^_^
979933
I'm giving u the program (in C), just go through it, then write the algo. It'll be easier.\n\n// Program :  MERGE  SORT\n\n#include<stdio.h>\n#include<conio.h>\n\nvoid mergesort(int *,int);\n\nvoid main()\n{\nint x[30],n,i;\nclrscr();\nprintf("Enter total no. of elements");\nscanf("%d",&n);\nprintf("Enter %d nos.",n);\nfor(i=0;i<n;i++)\n   scanf("%d",&x[i]);\n\nmergesort(x,n);\n\nprintf("After sorting in descending order :&#92;n");\nfor(i=n-1;i>=0;i--)\n  printf(" %d",x[i]);\n\ngetch();\n}\n\n\n\nvoid mergesort(int x[],int n)\n{\n  int aux[30],l1,l2,u1,u2,i,j,k,size;\n  size=1;   // merges file of size 1\n  while(size<n)\n  {\n    l1=0;   // initialize lower-bound of 1st. file\n    k=0;    // k is index for auxilliary array\n    while(l1+size<n)\n    {\n      // initialize remaining parameters...\n      l2=l1+size;\n      u1=l2-1;\n      u2=(l2+size-1)<n ? (l2+size-1):(n-1);\n      for(i=l1,j=l2;i<=u1 && j<=u2;k++)\n if(x[i]<=x[j])\n   aux[k]=x[i++];\n else\n   aux[k]=x[j++];\n      while(i<=u1)\n      aux[k++]=x[i++];\n      while(j<=u2)\n      aux[k++]=x[j++];\n\n      l1=u2+1;\n    } /* end of 2nd. while-loop */\n    for(i=l1;k<n;i++)\n       aux[k++]=x[i];\n    for(i=0;i<n;i++)\n       x[i]=aux[i];\n\n    size*=2;\n  } /* end of 1st. while-loop */\n}
1108238
I can provide you complete programs of data structures (linked list, array handling, stacks, queues, tress, graphs, and other algorithms). for this you have to provide me your information and details for what purpose you have to use this? and what your are doing currently. you can communicate me on my email address. abhi_patni@yahoo.com.
894119
Well... the easiest and most general approach is to  traverse the tree in preorder depth-first manner, writing the nodes to the file, and inserting a special separator (a centinel) at the end of each branch - just before the traversal process is about to move up:\n\nvoid write_node (struct node *root, FILE *out)\n{\n    root ? fprintf (out, "%s&#92;n", node_to_string (root, buffer)) :\n    fprintf (out, "%c&#92;n", SENTINEL);\n}\n       \nvoid read_tree (struct node *root, FILE *out)\n{\n   write_node (root);\n    if (!root) return;     \n    save_tree (root->left, out);\n    save_tree (root->right, out);\n}\n\nthen you can read this structure back in  in the same order:\nstruct node* read_node (FILE *in)\n{\n    char buffer[1024];\n    if (!fgets(in, buffer, 1024) || buffer[0] == SENTINEL) return 0;\n   return parse_node (buffer);\n}\n    \nstruct node *read_tree (FILE *out)\n{\n    struct node *root = read_node (out);\n    if (!root) return 0;\n    root->left = read_tree (out);\n    root->right= read_tree(out);\n}
737262
Imagine it this way... Today you have a project to prepare and for this you must go to a bookstore and find the information you need for it. This information usually doesn't come in one package, so in most cases you must join information you get from several "books" to be able to make your project. When you've done with it, another project comes along and you have to go through the entire process all over again. Of course nobody assures you that you'll be working at only one project independently... So there comes the time when you need to rearrange those "books" as they were before you used them, to be more easier and more quicker to search for them again next time.\nSame thing is with searching required files for completing a task which you are giving to the operating system.