number the nodes in the tree like this:
root=1
left child of root=2
right child of root=3
left child of {left child of root}=4
right child of {left child of root}=5
left child of {right child of root}=6
right child of {right child of root}=7
something similar to how you create a heap out of an array of elements.
and then visit each node write out its children to an array where element
numbered i is wriiten to array[i].
serialize thw array.
while deserializing it follow the steps to create a heap out of an array.
serialization takes O(n) time and O(n) space and so does deserialization.
On Mon, Jul 5, 2010 at 9:01 AM, harit agarwal <agarwalha...@gmail.com>wrote:
> in my approach
> a( b,c)-> this implies that nodes within parenthesis are child of a where b
> is left child and c is right child
> if there is no left child then we can use a(,c) and
> if there is no right child then use a(b) and
> if there is no child use only a
>
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