node *root;
node * build(int height)
{
node *r, *temp;
for(i=1, r=null; i<=height && root; i++)
{
temp = root;
root=root->left;
temp->right=r;
r=temp;
r->left = build(i-1);
}
return r;
}
void main()
{
node *tree;
//root contains original tree
tree = buld(n); // n=height of of new tree
}
i will explain with an example
ake an example 6-5-4-3-2-1
here N=6, height of the new tree = 3
so main will call build(3)
it will build the tree as follows
in build(3)
i=1 6 root 5-4-3-2-1
i=2 5 root 4-3-2-1
/ \
build(1) 6
in build(1)
i=1 4 root 3-2-1
so 5
/ \
4 6
i=3 3 root 2-1
/ \
build(2) 5
/ \
4 6
in build(2)
i=1 2 root 1
i =2 1 root null
/ \
build(1) 2
in build(1)
i=1 as root null it will be null
so 1
\
2
so 3
/ \
1 5
\ / \
2 4 6
this way it will work..
by the way can you give me an example that does not work ? it will be
great help as i can find the bug in my program. i am waiting for your
reply. for ne clarification i am here, you can ask me :)
On 3/17/06, pramod <[EMAIL PROTECTED]> wrote:
>
> Malay, I think your solution gives wrong results. Can you please verify
> and explain us.
>
>
>
>
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