@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 [email protected].
To unsubscribe from this group, send email to
[email protected].
For more options, visit this group at
http://groups.google.com/group/algogeeks?hl=en.
