Yes, you are right.
How about this:
bool isBST(root *t, int min=minint, int max=maxint)
{
return !t || ((t->value >= min) && (t->value <= max) &&
isBst(t->left, min, t->value) && isBst(t->right, t->value,
max));
}
On Nov 5, 11:04 pm, atul anand wrote:
> @Don : your algo wont work f
@Don,
Your method fails for the case below:
//Check if a binary tree is Binary Search Tree or not.
#include
typedef struct node
{
int value;
struct node *left,*right;
}nodeptr;
nodeptr *newnode(int x)
{
nodeptr* tmp = (nodeptr*)malloc(sizeof(nodeptr));
tmp->value=x;
tmp->left=N
@Don : your algo wont work for following tree :-
30
/ \
20 31
/ \
9 41
above tree is not a BST bcozz here 41 should lie on the right side of the
30but it is not.
so we need to keep track of max and min as we move left or right part of
the tree.and each node should
Understood, thanks.
On Tue, Nov 6, 2012 at 2:35 AM, Don 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
>
>
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 wrote:
> That would work. But a simpler approach is:
>
> b
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 wrote:
