It is given graph. It isn't ordinary graph, it is ladder. We say that our ladder has order $n$, because of number of nodes. Look at picture:
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My problem is: Let's start DFS on node number $1$. How many different DFS tree exist?

From my point of view I should count number of possible adjency lists. For example, for node number $1$ we have to possible adjency lists: $[2, 3]; [3, 2]$.

Is my approach ok ? What's your opinion ?

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You are close. Also remember the order of non-tree edges does not matter.

There are actually 2 cases for node 1:

  1. $$1 \to 2 \to 4 \to \dotsb$$ Start from node 4, it's just a DFS traversal of ladder of order $n-1$ rooted at node 4.
  2. $$1 \to 3 \to \dotsb $$ Start from node 3, it's just a DFS traversal of ladder of order $n-1$ rooted at node 3, with one extra branch from 4 to 2. And there are 2 ways to insert branch 2, since node 4 in ladder of order $n-1$ rooted at node 3 is of degree 2, having exactly 1 parent and 1 child.
  • $\begingroup$ So, from first case: $$T_{n-1}$$ from second case: $$T_{n-1}$$, $$T_{1}=1$$, Then $$T_{n} = 2T_{n-1} = 2^{n-1}$$. Ok ? $\endgroup$ – M.Swe Sep 22 '15 at 12:40

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