CareerCup

int printPath(NODE* root, int v)
   {
       if(root == NULL) return 0;
       if(root->v == v) 
       {
           printf("%d<-", root->v);
           return 1;
       }
       if( printPath(root->left,v) || printPath(root->right, v))
       {
            printf("%d<-", root->v);
            return 1;
       }
       else
       {
           return 0;
        }
    }

int search_path(node *root,int d)
{
if(!root) return 0;
if(root->data==d)
{
printf("%d",d);
return 1;
}
if(search_path(root->left))
{
printf("%d",root->left->data);
return 1;
}
if(search_path(root->right))
{
printf("%d",root->right->data);
return 1;
}
}

Do a BFS, maintain a map, before pushing the child of every node to queue, push the node and its parent to map. Once the node is found, use the map to find the parent and push the parent in stack till NULL parent is found, at that time pop out the elements of stack and print them

Do a BFS, maintain a map, before pushing the child of every node to queue, push the node and its parent to map. Once the node is found, use the map to find the parent and push the parent in stack till NULL parent is found, at that time pop out the elements of stack and print them

int search_path(node *root,int d)
{
if(!root) return 0;
if(root->data==d)
{
printf("%d",d);
return 1;
}
if(search_path(root->left))
{
printf("%d",root->left->data);
return 1;
}
if(search_path(root->right))
{
printf("%d",root->right->data);
return 1;
}
}

the question is to print route sequentially, so the correct one would be;

print_route(node * N, int key)
{

the question is to print route sequentially, so the correct one would be;

print_route(node * N, int key)
{
if(N==null)
exit(0);
if(key< N->value)
{
insert(L,n->value);
N=N->lchild;
print_route(N,key);
}
else if(key>N-> value)
{
insert(L,n->value);
N=N->rchild;
print_route(N,key);

}
else if(key==N->value)
{ insert(L,n->value);
}
while (L!=null)
{
printf("%d",L->info)
L=L->next;
}
}

insert(list* L, int k)
{
node=(list*)malloc(sizeof(struct list));

if(L==null)
{ L=node;
}
else if(node->info<L->info)
{
node->next=L;
L=node;
}
else
{
list* first=L;
list*nxtptr;
while (node->info > first->info)
{
nxtptr=first;
first=first->next;
}
nxtptr->next=node;
node->next=first;
}
return L;
}