@Decipher - something like this: void build_tree(Node*& node, int& index) { node = new Node(value[index++]); // create an empty node if (label[index] == 'L') return; // we are at leaf, there is no need to process further current subtree build_tree(node->left, index); // build left and right subtree build_tree(node->right, index); }
here we have arrays value - preorder traversal, and label array - label of the node (either L or N). root = NULL; index = 0; build_tree(root, index); -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.