Understood, thanks. On Tue, Nov 6, 2012 at 2:35 AM, Don <[email protected]> wrote:
> In English, that is > > A null tree is a binary tree. > Otherwise, it's a binary tree if the root value is greater than the > left child and less than the right child, and the left and right > subtrees are binary trees. > > Don > > On Nov 5, 2:48 pm, Don <[email protected]> wrote: > > That would work. But a simpler approach is: > > > > bool isBinTree(root *t) > > { > > return (!t) || ((!t->left || (t->value > t->left->value)) && > > (!t->right || (t->value < t->right->value)) && > > isBinTree(t->left) && isBinTree(t->right)); > > > > } > > > > On Nov 5, 2:04 pm, shady <[email protected]> wrote: > > > > > > > > > > > > > > > > > Hi, > > > Can we check this by just doing an inorder traversal, and then > checking if > > > it is in increasing order or not ? > > -- > 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. > > -- 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.
