Is there a way to convert from a general tree to binary SEARCH tree?

I think you just have to traverse the initial tree and insert each node into a binary search tree. After that, you will have converted your initial tree into a BST.\n\nFor traversing the tree click here\n\nFor binary search tree info and insertion method click here",

**\n\n\n This is an algorithm I created that can convert a tree data structure\n into a 2 degree tree true (Binary tree). Following is the node\n structure for the tree nodes.\n\n\n**\n\ntemplate<class T>\nclass Node\n{\npublic:\n T _data; // The Node Data\n\n std::vector<Node<T>*> _link; // Possible OutLinks from the Node.\n\n Node(){}\n\n void setValue(T D){ _data = D; }\n\n T getValue()const{ return _data; }\n\n ~Node(){}\n};\n\nFollowing is the tree adapter.\n\ntemplate <class T> \n\nclass tree{\n\n Node<T>*_root; // The Pointer holding the root node.\n\n int _degree;\n\n std::string _type; // shows the tree type,Binary Or Any other outdegree \n\n//degrees.\n\npublic:\n\n tree(){ _degree = 0; _root = NULL; }\n\n tree(Node<T>*R, int D) :_root(R),_degree(D){}\n\n\n Node<T>* getRoot()const{ return _root; }\n\n ~tree(){ _root = NULL; }\n};\n\n//.........\n//This is the Algorithm for converting a x-order tree to a binary tree.\n\n///*This Template function is used to convert a general tree into a binary tree \n\nwithout loosing any nodes,with the same root node.*/\n\ntemplate<class T> \n\ntree<T> makeBinaryTree(tree<T> _tree){\n\n Node<T>* node = _tree.getRoot();\n\n Node<T>* root = new Node<T>;\n\n root = node;\n\n\n int i = 0;\n\n int k = 0;\n\n std::queue<Node<T>*> que; // que used to save the links other than the \n\nleftmost one.\n\n std::stack<Node<T>*> s1; // stack for saving the tree nodes,while going deep \ninto left\n\n std::stack<char>s2;\n\n Node<T>* s3;\n\n char flag = ' ';\n\n while (true){\n\n while (root&&flag!='A'){\n\n s1.push(root);\n\n if (root->_link[0] == NULL && root->_link.size())\n\n s2.push('C');\n\n else if (root->_link.size() > 1){\n\n s2.push('B');\n\n }\n\n else\n\n s2.push('A');\n\n root = root->_link[0];\n\n }\n\n if (s1.empty())break;\n\n root = s1.pop();\n\n flag = s2.pop();\n\n\n if (flag == 'C'){ // handles the deep left node with any number of nodes \n\nother than in socket 0.\n\n while (true){\n\n i = 1;\n\n if (root->_link[0] == NULL&&root->_link.size() == 0){ flag = 'A'; break; \n\n}\n if (root->_link.size() >= 1){\n\n while (true)\n\n {\n\n if (root->_link[i]){\n\n root->_link[0] = root->_link[i];\n\n root->_link.erase(i);\n\n if (root->_link.size() > 1){\n\n s1.push(root);\n\n s2.push('B');\n\n }\n\n break;\n\n }\n\n ++i;\n\n }\n\n root = root->_link[0];\n\n }\n\n }\n\n }\n\n\n if (flag == 'B'){ // any node except the deep left node that has more links \n\nfrom it.\n\n i = root->_link.size()-1;\n\n k = i-1;\n\n while (K!=0){\n\n root->_link.at(k)->_link.push(root->_link.at(k + 1));\n\n --k;\n\n }\n\n s3->_link[1] = root->_link[1];\n\n root->_link.erase[1];\n\n s1.push(root);\n\n s2.push('A');\n // Now You have to manage link 1 of s3.\n\n s3 = s3->_link[1];\n\n\n\n if (s3->_link.size()){\n\n //TODO...\n\n //Testing...\n\n root = s3;\n\n }\n\n //AT the end \n\n s3 = NULL;\n\n\n\n }\n\n\n\n\n\n if (flag == 'A'){ // the safe nodes,i.e having only one node from it \n\n,other than the deep left node\n\n s3 = root;\n\n }\n\n\n\n }//end of main while loop.\n\nreturn (new tree<T>(node,2));\n\n}\n",