I need a function that must verify if a binary tree is a searching binary tree.help :)?
965114
I think you mean that you want to verify that your search is a binary search? Because binary trees don't search. \n\nIn order to verify the correctness of the search, print out the memory contents at each iteration of your search and review the results.
228645
Have you tried getting the object as a FileInfo or something like that and then checking the type based on that?\nThats the best I can think of.
836395
Ref topnode = call readnode.\n\nFunction readnode is:\nGet an element from the file.\nIf the element is a special code (e.g. -1) return NULL.\nConstruct refnode.\nSet refnode's left leaf = readnode(refnode).\nSet refnode's right leaf = readnode(refnode).\nReturn refnode.\nEnd function\n\nTo use this loading function, write the file depth first, recursively saving the left node then the right node.  If a child node doesn't exist (e.g. you're at the bottom of the tree), write the special code.
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).
67914
hey!! looking for ur c++ project.. source code online?? isnt a good idea!\n\nwell thetopic u've selected is idiotic.. \nu better submit an original work.. or else u'll loose marks!\n\nn if programming is ur problem.. n u're lookin for an easy way out.. try looking for tutorial sites for c++.. n search for samples .. but then be assured that ur teacher will come to know aboiut the authenticity just by the format.. \n\nwell if u're smart enough to edit the stuff n then submit it.. only then its of some use.\n\nn by the way.. i've heard of this project .. its common.. i mean even some of my batchmates did this project.. a very ordinary 1... so u may get some help from ur seniors as welll!!
879190
it depends on your programming language! does it do xpath? do you have to parse it? is it in a pre-parsed tree???
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
1159892
Sequential search:\n\nhttp://cprogramming.com/discussionarticles/sorting_and_searching.html
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.
838865
What language is this?  It looks like the result of a night of unbridled passion between C, C++, and Java!\n\nSo, you know that "x->y" is equivalent to "(*x).y", right? So, even though you never used . directly, it's apparently what the compiler is seeing. Of course, . can only be used with a struct, and it is complaining that you are using it with something that is not a struct. However, the error is confusing, since the problem is really caused by a problem with pointers.\n\n Consider b[10]  (and b[9]).  b is a pointer to blocks. The [ ] operator dereferences this pointer, so that b[10] has type block, not block *.  Since -> can only be used with pointers (block *), the compiler generates an error. \n\nThink carefully about what you want to do, draw a picture on paper, and make sure your types all make sense.  Good luck.
240615
I am giving the solution in C++ using recursion..\n\n!!\n\n#include<iostream.h>\nvoid printstars(int n)\n{\nfor(int k=0;k<n;k++)\n     cout<<"*";\ncout<<endl;\n}\nvoid print1(int n)\n{\nif(n<=1)\n   printstars(n);\nelse\n   printstars(n-2);\n}\nvoid print2(int m,int n)\n{\nif(m==n)\n   cout<<endl;\nelse \n{\n   printstars(m);\n   print2(m+2,n);\n}\n}\n\nmain()\n{\nint n,m=1;\ncout<<"Enter an odd number n";\ncin>>n;\nprint2(m,n);\nprint1(n);\n}\n\nThis produces the output u needed and also uses recursion.
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You can do this by hand.\nCall your number x.\nWrite the binary digits of x as b_i :\nx = sum_n=0_n=N b_n.2^n\n\nThe algorithm is simple: compare your number x to the largest power of two (2^N), subtract if necessary, and decrease n until finished (n=0).\n\nStart with n=N , the largest integer such that X >=2^n\nIf x >= 2^n, then {\n b_n = 1;\n x = x-2^n;\n n = n-1; \n} else {\n b_n 0;\n}\n\nThe above is pseudocode, you could do it in C, PERL, Python, Java etc.\n\nSo: x= 121 = 64+57=(64+32)+25=(64+32+16)+9\n=(64+32+16+8)+1 =64+32+16+8+0+0+1\n\nThus 121 (base 10) = 1111001 (base 2)
423398
here...I'll tell you what to write and you write it...\n\n/* outline the program */\n\n// void main()...etc..\n\n/* define your variables */\n/* you'll need a pair of char pointers */\n/* and some numbers to hold converted values */\n\n\n/* input the first binary as a string */\n\n/* input the second binary as a string */\n\n/* convert them to decimal */\n\n/* add them */\n\n/* fprintf them */\n\n/* end program */\n\n\nthat's all.\nI write all my programs in commented logic first then fill in the blanks so that it does what I meant.  When I'm done, the program follows the flow and is already commented for the next person to follow.
161686
Easy if you restrict to integer types.\n\nCreate the following macro:\n\n#define IS_SIGNED(type) (((type) 0xFFFFFFFF) < 0)\n\nYou can now use it direction in your code as such:\n\nif (IS_SIGNED (char))\n{\n   stuff if it is signed\n}\nelse\n{\n   stuff if it is unsigned\n}\n\nNote that this implementation assumes that you maximum integer type is 32 bits. Pretty easy to change if it is otherwise though.\n\nI have done a LOT of multiple-platform code, char signing and byte ordering are the two biggest headaches.\n\nHope this helps.
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Most search engines follow the rules and symbols for Boolean logic.  If you spend anytime using search engines you might find the following article very informative.
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#include <stdio.h>\n\nint main(void) {\n\nint i, j, count; //C Sucks, wait till you take java\n\nfor(i=1; i<=1000; i++){&#92;&#92;goes through all numbers 1 to 1000\n    count=0;\n    for(j=2; j<i; j++)&#92;&#92;goes through all numbers 2 to i\n    {\n     if(i%j==0)&#92;&#92;checks to see if it i is evenly div. by i\n        count++;\n    }//nested for\n    if (count ==0)\n        printf("%d&#92;n", i);\n}//for\n\nreturn 0;\n}//main\n\n\n//Hope that helps!
966120
try this stuff... pls understand the logic...\n\n#include<stdio.h>\n\nint main()\n{\nint no,rem;\nscanf("%d",&no);\nwhile (no)\n{\nrem = no % 10;\nprintf("%d",rem);\nno = no / 10;\n}\n}