I am not getting what you said. For given tree

a(2)b(7)c(5)d(2)e(6)f(9)g(5)h(11)i(4)   if i say given node is 6 and
distance is 1 then the output should be 7,5,11.

The nodes below the given nodes can be easily printed. I am not
getting the way to print nodes above the given node

On Oct 31, 1:30 pm, SAMMM <[email protected]> wrote:
> As the problem states that the distance can be upwards and downwards .
> So we considering both the case . I am going to implement BFS to
> implement it (Because requires the poped elements to trace back to the
> Source node to check whether path is sorted)
>
> Consider the tree Given in this link .. I am taking tht as reference 
> .http://en.wikipedia.org/wiki/Binary_tree
>
> Suppose the Source node is the left node of the root node . naming the
> nodes from A to I breadthwise .
>
> a(2)b(7)c(5)d(2)e(6)f(9)g(5)h(11)i(4)
>
> Case 1:- Traving from the root we note the distance from the Root to
> the source node but don't push it in the queue.
>
> Source node - b , Distance =3
> -------------------------------------------------------------------
> Queue    = b | d e | g h
> Distance = 0 | 1 1 | 2 2
> PrevNode = --| b b | e e
>
> So we don't get any distance as 3 . Leave this and reinitialize the
> Queue , Distance and PrevNode .
>
> Case 2 :-Any now Repeat this process from Root node to the source node
> to see whether we get the Distance as 3 .
>
> Case 3:- If the distance from the Root to the Source node is less than
> 'K' distance , then repeat the BFS process from the Root .
>
> Source node - b , Distance =3
> -------------------------------------------------------------------
> Queue    = a | c | f | I
> Distance = 0 | 1 | 2 | 3
> PrevNode = --| a | c | f
>
> Now the distance from the Root to the source node(d) is 1 here , s o
> we need to search for the node whose distance from the root is (K-d) .
>
> Here we will find the path with K=3 distance from the source node ,
> But it will not be valid because it is not sorted .. So no node
> present will be printed .
>
> Hope I am clear .

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