just do one thing print inorder. if sorted it is a BST
On Jun 19, 2:57 pm, divya <sweetdivya....@gmail.com> wrote:
> i have found the following code on net to check if the binarytreeis
> bst or not
> /*
> Returns true if a binarytreeis a binary searchtree.
> */
> int isBST(struct node* node) {
> if (node==NULL) return(true);
> // false if the min of the left is > than us
> if (node->left!=NULL && minValue(node->left) > node->data)
> return(false);
> // false if the max of the right is <= than us
> if (node->right!=NULL && maxValue(node->right) <= node->data)
> return(false);
> // false if, recursively, the left or right is not a BST
> if (!isBST(node->left) || !isBST(node->right))
> return(false);
> // passing all that, it's a BST
> return(true);
>
> }
>
> int minValue(struct node* node) {
> struct node* current = node;
> // loop down to find the leftmost leaf
> while (current->left != NULL) {
> current = current->left;}
>
> return(current->data);
>
> }
>
> and maxvalue also similarly returns right most vaslue oftree..
>
> now i have a doubt that here v r comparing the node->data with only
> the min node nd max node.. shd nt we compare the node data with its
> left node and right node only.
> as it can b a case that node value is greater than min but less than
> its left node.. nd here we r nt checking that...
>
> plzzzzzz correct me if i m wrong...........
>
> and is there any other efficient method to find isbst?
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