binary tree java implementation ?
89402
Code:\npublic class BTNode \n{\n    private Object data;\n    private BTNode left, right;\n    public BTNode(Object initialData, BTNode initialLeft, BTNode initialRight)\n        { \n            data = initialData;\n            left = initialLeft;\n            right = initialRight;\n        }\n    public BTNode()\n            { this(null, null, null); }\n    \n    public BTNode(Object newdata)\n            { this(newdata, null, null); }\n    \n    //NODE methods\n    public Object getData()                     //return and set data in node\n            {return data;}\n    \n    public void setData(Object data)\n            {this.data = data; }\n    \n    public BTNode getLeft()                 //return and set left child\n            {return left; }\n    \n    public void setLeft(BTNode left)\n            {this.left = left; }\n    \n    public BTNode getRight()                //return and set right child\n            {return right; }\n    \n    public void setRight(BTNode right)\n            {this.right = right; }\n    \n    public boolean isLeaf()                 //return true if this is leaf node\n            {return (left == null)&&(right == null); }\n \n// TREE Methods - for tree starting at this node\n// get specific data from a tree\n    \n        public Object getLeftmostData()\n        { if(left == null)\n               return data;\n          else \n              return left.getLeftmostData(); }\n        \n        public Object getRightmostData()\n              { if(right == null)\n               return data;\n          else \n              return right.getRightmostData(); }\n// print data in a tree\n        public void inorderPrint()\n                {\n                    if (left != null)\n                        left.inorderPrint( );\n                    System.out.print(data+"&#92;t");\n                    if (right != null)\n                        right.inorderPrint( ); }\n        \n        public void preorderPrint()\n                {\n                    System.out.print(data+"&#92;t");\n                    if (left != null)\n                        left.preorderPrint( );\n                    if (right != null)\n                        right.preorderPrint( ); }\n        \n        public void postorderPrint()\n                {\n                    if (left != null)\n                        left.postorderPrint( );\n                    if (right != null)\n                        right.postorderPrint( );\n                    System.out.print(data+"&#92;t"); }\n        \n// pretty print a tree\n        //public void print(int depth);\n \n//Interface for Nodes in Binary Tree (3)\n//removing nodes from a tree\n//returns reference to tree after removal of node\n        public BTNode removeLeftmost()\n              {\n                    if (left == null)                               // we are as deep as we can get\n                        return right;                               // remove this node by returning right subt ree\n                    else\n                        {                                           // keep going down recursively\n                            left = left.removeLeftmost( );\n                                                                    // when done, return node that act ivated this\n                                                                    // method as root of t ree\n                            return this; }}\n        \n        public BTNode removeRightmost()\n        {\n                    if (right == null)                               // we are as deep as we can get\n                        return left;                               // remove this node by returning left subt ree\n                    else\n                        {                                           // keep going down recursively\n                            right = right.removeRightmost( );\n
293992
Use TOBA (see link below)\n\nToba translates Java class files into C source code. This allows the construction of directly executable programs that avoid the overhead of interpretation.
256487
in here www.thefreecountry.com/compilers/pascal.shtml
780385
Try it out - you need a semicolon at the end of the first print statement for it to compile.\nIn java,\nit prints out:\n0000000000\n0123456789\n0246802468\n0369258147\n0482604826\n0505050505\n0628406284\n0741852963\n0864208642\n0987654321
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Object\n\n1) enclose it in braces\nString [] apple = {"a", "b", "c"};\n\n2) read it into what?  Array?  ArrayList?  I can do it in four lines...\nString line;\nBufferedReader br = new BufferedReader(new FileReader("build.xml"));\nwhile ((line = br.readLine()) != null) {/*do something here*/}\nbr.close();\n\n3) string.replace\nString banana = "This'is'a'test";\nSystem.out.println(banana.replace("'", ""));\n\n(In case that's hard to read - the method is "replace" openparin doublequote singlequote doublequote comma doublequote doublequote closeparin)
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
9135
Since the keys in a B-tree is sorted, just keep following the first (leftmost) subtree until a node has no leftmost subtree.  Then the smallest element in that node will be the minimum key.
408935
Hey,\nthe code in JAVA would look like this ::\n\nclass test {\npublic static void main(String a[]) {\nfor(int i=1;i<100;i++) {\nSystem.out.print("&#92;t"+i+"&#92;n");\n}\n}\n}
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}
826904
for (int i = 0; i < arrayA.length; i++)\nfor (int j = 0; j < arrayA[0].length; j++)\n  arrayC[i][j] = arrayA[i][j] + arrayB[i][j];\n\nSystem.out.println("  Values for arrayC");\nfor (int i = 0; i < arrayC.length; i++)\n   for (int j = 0; j < arrayC[0].length; j++)\n   {\n      System.out.print(arrayC[i][j]);\n      if(j < arrayC[0].length - 1)\n        System.out.print("   ");\n      else\n        System.out.println("");\n  }
927748
You first define an array NodesList with index 0 as origin (C standard) containing all your nodes. The variable NodesNumber is the number of nodes. The key element is to define the operators S (for successor) and P (for predecessor) that will make the array behave as a circular list.\nS(NodesList[ i ]) = NodesList[ i + 1] if i < NodesNumber - 2\nand for the last node\nS(NodesList[ NodesNumber - 1 ]) = NodesList[ 0 ] \nSimilarly for P\nP(NodesList[ i ]) = NodesList[ i - 1] if i > 0\nand for the first node (origin)\nP(NodesList[ 0 ]) = NodesList[  NodesNumber - 1 ] \nThe algorithm has 2 parts\n// forward\nStartNode := NodesList[ NodesNumber - N  ]\nfor I := 1 to N\n  begin\n     StartNode := S(StartNode);\n     Node := StartNode;\n     Path := StartNode;\n     for J := 1 to N - 1\n        begin\n           Node := S(Node);\n           Add Node to Path \n        end \n      Record Path\n  end;\n// backward\nStartNode := NodesList[ N  ]\nfor I := 1 to N\n  begin\n     StartNode := P(StartNode);\n     Node := StartNode;\n     Path := StartNode;\n     for J := 1 to N - 1\n        begin\n           Node := P(Node);\n           Add Node to Path \n        end \n      Record Path\n  end;\nIn the second part the starting point is changed and S() are changed by P()
440005
Pseudocode, I'm not up on my Java syntax:\n\nint sum=0;\nfor(int i=0;i<array.size;i++)\n  sum+=array[i];
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If we don't plan on modifying a search tree, and we know exactly how often each item will be accessed, we can construct an optimal binary search tree, which is a search tree where the average cost of looking up an item (the expected search cost) is minimized.\n\nAssume that we know the elements and that for each element, we know the proportion of future lookups which will be looking for that element. We can then use a dynamic programming solution, detailed in section 15.5 of Introduction to Algorithms, to construct the tree with the least possible expected search cost.\n\nEven if we only have estimates of the search costs, such a system can considerably speed up lookups on average. For example, if you have a BST of English words used in a spell checker, you might balance the tree based on word frequency in text corpuses, placing words like "the" near the root and words like "agerasia" near the leaves. Such a tree might be compared with Huffman trees, which similarly seek to place frequently-used items near the root in order to produce a dense information encoding; however, Huffman trees only store data elements in leaves.
435211
Color tree (or Munsell color system)
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http://www.thefreecountry.com/compilers/cpp.shtml
414394
Try \n1) Let Us C++\n2) Core Java 2.0 Complete reference\nRead from the very first page and do practical examples page by page. You will defintely able to learn just like me.
1155261
You can do Binary Searching in two ways \n\n1. Using loops.\n2.Using recurssion.\n\nWhat even,Lets now discuss how it works.\n\nIt works only in a sorted array or in a sorted sequence.\n\nWe are now going to divide the array into two parts using a median or a middle value.Before starting we should need to have a lower bound and an upper bound.\n\nIn starting the lower bound value would be zero and the upper bound value should be no . of elements in the array.\nmedian is always (lowerbound+upperbound)/2\n\n\n\nLet us set the 'median' as mid and the number you want to search as 'num'\n\nAt the first step check mid=num\nIf so the value is found\n\nelse\n\ncheck num<mid\nIf num<mid\nset upperbound = mid\nand find new mid as\n(lowerbound+upperbound)/2\n\nElse if,\nnum> mid\nset lower bound=mid\nand find the new mid value,\nas \nmid=(lowerbound+upperbound)/2\n\n\nAnd contine the process until the value is found or the lowerbound=upperbound(Means no element in the array)\n\n\nI have shown only the concept.\nYou can do it by using loops,Or by using recurrsion.You can use Divide and Conqure method too(recursive  method).