@harshit can you provide  a running code and prove how its running
complexity is better than the complexity of my code

On Sat, Aug 6, 2011 at 6:44 PM, harshit sethi <[email protected]>wrote:

> inordersucc(struct tree* p)
> {
> if(p->right !=NULL)
> return p->right;
>
> struct tree* ptr,*psuc;
> ptr=root;
>
> while(ptr->info!=x->info)
> {
>
> if(x->info<ptr->info)
> {
> psuc=ptr;
> ptr=ptr->left;
> }
>
> else
> ptr=ptr->right;
> }
>
>
> return psuc;
> }
>
> On 8/6/11, Anuj Kumar <[email protected]> wrote:
> > i am sending the running code :)
> >
> > #include<iostream>
> >
> > using namespace std;
> > struct node
> > {
> > int data;
> > struct node *lc,*rc;
> > node(int d)
> > {
> > data=d;lc=rc=NULL;
> > }
> > };
> > int ret,fl;
> > void inorder(struct node *root,int d)
> > {
> > if(ret==1)return;
> > if(root==NULL)return;
> > inorder(root->lc,d); if(ret==1)return;
> > if(fl==1){cout<<root->data;ret=1;}
> > if(root->data==d){fl=1;}
> > inorder(root->rc,d); if(ret==1)return;
> > }
> > int main()
> > {
> > struct node *root=new node(2);
> > root->lc=new node(1);
> > root->rc=new node(3);
> > root->rc->rc=new node(6);
> > ret=0;fl=0;
> > inorder(root,3);
> > return 0;
> > }
> >
> > On Sat, Aug 6, 2011 at 5:50 PM, Aman Goyal <[email protected]>
> wrote:
> >
> >> .. pseudo code for finding inorder successor of a node without parent
> >> field..
> >>
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> >
> >
> >
> > --
> > Anuj Kumar
> > Third Year Undergraduate,
> > Dept. of Computer Science and Engineering
> > NIT Durgapur
> >
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>


-- 
Anuj Kumar
Third Year Undergraduate,
Dept. of Computer Science and Engineering
NIT Durgapur

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