thing is that u r using recursion and we don't have to use it(
recussion use memory indirectly) as per the question
On Thu, Apr 29, 2010 at 3:55 PM, Algoose Chase wrote:
> If you mean to convert the binary tree to binary search tree directly , then
>
> BinarytoBST(Node* root)
> {
> if( root =
@ashish
i forgot recussion uses memory but if we have to do it without using
stack also then
pickup the root and add it to the bst and to fill the vacant position
of root choose left node and make it root and to adjust previous right
node at it to leaf
eg :
D
/
If you mean to convert the binary tree to binary search tree directly , then
BinarytoBST(Node* root)
{
if( root == nulll) return;
BinarytoBST(root->left);
BinarytoBST(root->right);
if( root->left )
Node* NodeL = MAX(root->left);
if ( root->right )
Node* NodeR = MIN(root->ri
ya post order traversal will not have these problem theme time i
haven't thought the problem with pre and inorder.
On Wed, Apr 28, 2010 at 10:16 PM, Vivek S wrote:
> @Rajesh Patidar
> I think we should do in Post order traversal alone. If we go by
> Preorder/Inorder we might lose track of childre
Hi,
How do you define "without extra space" ?
Doing any order traversal either using recursion or using iteration is going
to take extra space .
If you are given a binary tree represented by pointers that points to
children nodes is it possible to do a heap sort without an array ?
On Thu, Apr 29
my choice is build a min heap .sort the array with heap sort.then find the
median of the sorted array and build tree
On Wed, Apr 28, 2010 at 10:16 PM, Vivek S wrote:
> @Rajesh Patidar
>
> I think we should do in Post order traversal alone. If we go by
> Preorder/Inorder we might lose track
@Rajesh Patidar
I think we should do in Post order traversal alone. If we go by
Preorder/Inorder we might lose track of children node that is currently
being inserted into the BST. - correct me if im wrong :)
On 28 April 2010 15:30, Rajesh Patidar wrote:
> pickup node in any order no matter(pre
@rajesh can u explain your soln
how u r doing inorder, pre or whatever (without using stack) and at same
time build BST
On Wed, Apr 28, 2010 at 3:30 PM, Rajesh Patidar wrote:
> pickup node in any order no matter(pre,post,inorder) and just one by
> one. start adding the node into bst no need to u
@ Rajesh: there may be a problem with this solution.
Suppose I start detaching the nodes from the binary tree in the following
order - > Root, Left, Right.
So as soon as i detach the root of the binary tree and form a new BST with
it ( on which i m going to make further node additions), I am left w
pickup node in any order no matter(pre,post,inorder) and just one by
one. start adding the node into bst no need to use extra space u have
to just ditach the node from binary tree and attach it in bst.
On Wed, Apr 28, 2010 at 1:18 AM, Ashish Mishra wrote:
> How to build BST from binary tree in p
On Wed, Apr 28, 2010 at 1:18 AM, Ashish Mishra wrote:
> How to build BST from binary tree in place i.e without extra space ??
>
Are you looking for:
http://discuss.joelonsoftware.com/default.asp?interview.11.781167.4
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