@ sanchit
m dong inorder traversal and at every step chking whether the node's p
pointer is pointing to its inorder predecesor, which is temp or not and
making it NULL otherwise. when count is 0 the node do not hv any predecessor
so m directly pointing that to NULL.
*please see the below alg0 , th
explain algo instead of writing the code.
thanx
On Mon, Feb 14, 2011 at 9:28 PM, jalaj jaiswal wrote:
> @tushar that would modify the tree structure
>
> here is a different approach
>
> int count=0; //global
> void modified_inorder(node *root){
>if(root!=NULL){
> modifie
@tushar that would modify the tree structure
here is a different approach
int count=0; //global
void modified_inorder(node *root){
if(root!=NULL){
modified_inorder(root->left);
node *temp;
if(count==0){
root->p=NULL;
I think the following algo should work:
1. Create a DLL of the inorder traversal of the tree
2. for each node, check whether P of that node points to the previous node
in the DLL or not.
3. If not, assign it value NULL
--
Tushar Bindal
Computer Engineering
Delhi College of Engineering
Mob: +919
What is meant by :
"preceding P’s node in the in-order
traversal of the tree" ?
On Mon, Feb 14, 2011 at 7:03 PM, bittu wrote:
> You have a tree, in which each node has an additional pointer, P. P
> can either be NULL or point a node preceding P’s node in the in-order
> traversal of the tree. Wr
You have a tree, in which each node has an additional pointer, P. P
can either be NULL or point a node preceding P’s node in the in-order
traversal of the tree. Write a program to check in a tree if each
node’s P is assigned correctly. If not, make P null
Thanks
Shashank
--
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