@Rahul : if you check the flow properly ,(lets concentrate on the root node
, call other as left and right subtree) you will find that after done with
reversing root->left->left subtree , it reaches root(backtrack to root)
node and then swap root->left and root->right.
now because it is inorder way of traversal , we are done with
function(&(*node)->Left); now next recursion for root node is
function(&(*node)->Right);.
function(&(*node)->Right); ---> this will again do reversing steps in
root->right subtree.(reverting back the old state).
after backtracking it reaches root.
but wait there is no swapping part after function(&(*node)->Right) , but it
is after function(&(*node)->Left);
but we have as discussed before function(&(*node)->Right) is the only
recursion root needs to complete in order move out of the function.
hence root->left and root->right remain swapped.
but program is similar to swapping root->left and root->right which can be
done in constant time but taken O(n) to do the same.
hope you have understood it :)
On Fri, Feb 10, 2012 at 10:16 AM, Rahul Menon <menonrahul1...@gmail.com>wrote:
> What about just the root being reversed?
> Why is it different only in case of root?
>
> Rahul
>
> On Feb 9, 10:52 pm, Don <dondod...@gmail.com> wrote:
> > Because it reverses one side twice and the other side not at all.
> > It does a lot of work to accomplish nothing.
> > Don
> >
> > On Feb 9, 9:06 am, Rahul Menon <menonrahul1...@gmail.com> wrote:
> >
> >
> >
> >
> >
> >
> >
> > > How come it just reversed the root? I still dont get it!
> >
> > > Rahul
> >
> > > On Feb 9, 7:57 pm, Don <dondod...@gmail.com> wrote:
> >
> > > > It appears to be an attempt to reverse the tree. However, there is a
> > > > problem. It reverses the left sub-tree, then swaps the left and right
> > > > sub-trees. Then it reverses the right sub-tree. But wait! The
> original
> > > > left sub-tree which we reversed is now the right sub-tree, so we
> > > > actually unreversed it. And the left sub-tree has never been reversed
> > > > at all. So I don't think that it will work. The actual result will be
> > > > that left and right will be swapped in the root node. Beyond that,
> > > > there will be no change.
> >
> > > > To make it work as intended, either do the two recursive calls one
> > > > after the other, or change the second one from Right to Left.
> >
> > > > Don
> >
> > > > On Feb 9, 8:38 am, Rahul Menon <menonrahul1...@gmail.com> wrote:
> >
> > > > > What does this function do?
> >
> > > > > void function(Node **node){
> > > > > if(*node!=NULL){
> > > > > function(&(*node)->Left);
> > > > > Node *temp;
> > > > > temp = (*node)->Left;
> > > > > (*node)->Left= (*node)->Right;
> > > > > (*node)->Right = temp;
> > > > > function(&(*node)->Right);
> > > > > }
> >
> > > > > }- Hide quoted text -
> >
> > > > > - Show quoted text -- Hide quoted text -
> >
> > > - Show quoted text -
>
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