convert bst to doubly linked list inorder and find middle element using slow
and fast pointer concept
On Tue, Sep 27, 2011 at 2:18 PM, Ankur Garg ankurga...@gmail.com wrote:
I was thinking of making the tree balanced with equal no of nodes in left
and right subtree
Median will be
I was thinking of making the tree balanced with equal no of nodes in left
and right subtree
Median will be root in this case
Regards
Ankur
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do the inorder traversal as soon as reached at n/2th element that will be
median.
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@Anshu
Will u be using extra space to store ur nodes in traversal ?
On Tue, Sep 27, 2011 at 2:22 PM, anshu mishra anshumishra6...@gmail.comwrote:
do the inorder traversal as soon as reached at n/2th element that will be
median.
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keep a global pointer and a global variable check=0
do inorder traversal..instead of printing the value do
if(check==0)
save pointer
check==1
else
check==0;
correct me if m wrong
On Tue, Sep 27, 2011 at 2:22 PM, anshu mishra anshumishra6...@gmail.comwrote:
do the
int bstMedian(node *root, int n, int x)
{
if (node-left) return bstMedian(root-left, n, x);
x++;
if (x == n/2) return node-val;
if (node-right) return bstMedian(root-right, n, x);
}
call bstMedian(root, n, 0);
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