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 <anuj.bhambh...@gmail.com> 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 <aman.goya...@gmail.com> 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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>
>
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