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I was asked to find an example of two non-isomorphic trees with the same amount of vertices and the same sequence of edges.

I can't seem to understand why it's not isomorphic.

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    Did you mean "nonisomorphic" (meaning not having the same form)? And also what would "same sequence of edges" mean? – 2017-02-03
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    That each tree has the same sequence of degrees..I hope I'm clear enoguh..sorry English isn't my first language – 2017-02-03
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    Two trees that are not isomorphic :), yes exactly what you mean – 2017-02-03
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    Do you mean that if one of the graphs has a vertex of degree 3 then the other graph must also have a corresponding vertex of degree 3? Like there should exists a bijective function $f:V_1 \to V_2$ from the vertices $V_1$ of graph 1 to the vertices of graph 2, $V_2$, such that $\text{deg}(f(v)) = \text{deg}(v)$? Or more plainly: The degrees of the vertices of one graph should "match" the degrees of the vertices of the other. Or am I misunderstanding you? – 2017-02-03
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    Yes! Your first statement matches the requirement :) – 2017-02-03
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    @Squid, by not isomorphic I mean that there isn't a bijective function between the two trees – 2017-02-03
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    Okay that's comparatively easy. You could actually just draw a tree with a few vertices like 6 and trial and error your way into drawing a new tree with the same number of vertices and thedegrees. A more elegant method would be to do something like this: Draw a sufficiently large tree. Now "detach" a "branch" of the tree and attach it somewhere else. This preserves number of vertices and most of the degrees while you just try make sure that what happens at the point of "cutting" and the point of "reattaching" cancel out. But just try this thing of moveing a branch and you'll see what I mean. – 2017-02-03
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    (For an isomorphism the bijection between the vertices should preserve edges in the sense that if $v_1$ and $v_2$ are connected, $v_1 \sim v_2$ then their images are connected too, $f(v_1) \sim f(v_2)$. My bijection just preserved degrees but not necessarily which elements are connected to which. Looks kinda similar but they're different definitions. I was just trying to avoid ambiguity by a formal definition. You don't really need to use it) – 2017-02-03
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    Thank you so much Squid! You should write it as an answer so I could thumb it up ^_^ – 2017-02-03
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    Draw two trees with degree sequence $3,3,2,1,1,1,1,$ one where the two vertices of degree $3$ are adjacent, and one where they are not adjacent. – 2017-02-04

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The most straightforward way is to list all the trees of with a certain number of vertices and go on until one finds two graphs which match the criterion. While tedious it does not take particularly long as there are for example just 2 trees with 4 vertices, 3 trees with 5 vertices, 6 trees with 6 vertices, 11 with 7 vertices, 23 with 8 vertices, and so on. (OEIS A000055) While the number of trees with a fixed number of vertices increases exponentially one encounters pair of examples before it grows out of control.

A more elegant way is using a conceptual idea I outlined in the comments where one considers ways to modify a preexisting tree in such a way that most of the trees structure, number of vertices and degrees of vertices are conserved.

One such operation is that of "cutting off a branch and reattaching it onto another branch". I will assume it to be visually clear what that would mean; a "branch" being a sort of sub-tree of the main tree. Only two vertices will have their degrees change; the vertex where the branch is cut and the vertex at where it is reattached, but the overall shape of the tree can be made to change significantly.

Interesting sub-problems to this problem is to find the smallest number of vertices you need to form two of these trees, or consider whether or not you can have two non isomorphic trees with the same degrees but which cannot be transformed into eachother by rearranging branches (or alternately where one rearrangement isn't enough)

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If you're looking for two non-isomorphic trees with the same degree sequence, I think the smallest example must be

    O                              O
    |                              |
O---O---O---O---O    and   O---O---O---O---O
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For orders $n \leq 5$, there are no non-isomorphic trees with the same degree sequence. For $n=6$, Henning Makholm's example is the unique case. For $n=7$, we have the following possibilities (with vertices marked as their vertex degrees):

Degree sequence $(3,3,2,1,1,1,1)$:

enter image description here

Degree sequence $(4,2,2,1,1,1,1)$:

enter image description here

Degree sequence $(3,2,2,2,1,1,1)$:

enter image description here

These were computed using geng which comes packaged with Nauty.