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 <dondod...@gmail.com> 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 <sinv...@gmail.com> wrote:
>
>
>
>
>
>
>
> > Hi,
> > Can we check this by just doing an inorder traversal, and then checking if
> > it is in increasing order or not ?
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