BST check function..whats wrong?

[roy.42]

/*Check if the variable root is a BST*/
113 int is_BST(tnode_t *node){
114     return( (node == NULL) || (
115                             (node->data > node->left->data) &&
116                             (node->data < node->right->data) &&
117                             (is_BST(node->left)) &&
118                             (is_BST(node->right))
119                             )
120         );
121 }

Im looking to return true if it is a BST, flase otherwise, im getting an "error: syntax error before '}' token"

aNy help greaty appreciated, im sure its just something simple, i just cant spot it..

EDIT::
Well i managed to get rid of the error, forgot to put a function declaration in..but now im getting segmentation faults when i run it.

[MacGyver]

Might want to check if node->left and node->right are NULL or not, although that isn't your problem currently.

I don't see the syntax error, and you weren't kind enough to tell me what line it was on, so I'm sorry.

Edit: See my first statement to fix the seg faults.

[roy.42]

oh sorry, it was on 121...
Oh, yeah, where abouts should i be checking them? Will i still be able to keep it in the one return statement do you thnk?

[MacGyver]

Yeah, I suppose you could, but I wouldn't recommend making giant return statements.

[roy.42]

Thanks for your help thus far, definately getting somewhere..

114 /*Check if the variable root is a BST*/
115 int is_BST(tnode_t *node){
116     if (node == NULL){
117         return 1;
118     }
119     else if( node->left!=NULL && node->right !=NULL){
120         if ((node->data > node->left->data) &&
121             (node->data < node->right->data) &&
122             (is_BST(node->left)) &&
123             (is_BST(node->right))
124             ){
125                     return 1;
126             }
127             else{
128                     return 0;
129             }
130     }
131 
132     else if ( node->left!=NULL && node->right == NULL){
133         if((node->data>node->left->data) &&
134             (is_BST(node->left))){
135                 return 1;
136             }
137             else{
138                 return 0;
139             }
140     }
141 
142     else if ( node->left==NULL && node->right != NULL){
143         if((node->data<node->right->data) &&
144             (is_BST(node->right))){
145                 return 1;
146             }
147             else{
148                 return 0;
149             }
150     }
151 else return 0;
152 }

Ok so i've split it up a bit..it compiles and runs, only problem is, its returning 0 for sometihng i know is a BST..

[ZuK]

Your code returns false if both left and right are NULL. Guess that would be a valid tree.
Kurt

[roy.42]

Hahahaha, thanks.
Oh well, to anyone else that was looking add these lines

119     else if (node->left == NULL && node->right == NULL){
120         return 1;
121     }