A tree is a connected 1-complex $X$ every two points $x,y$ of which can be separated by a point $z\neq x,y$. (We could let $X$ be a more general topological space here.)

Going one dimension up, call a connected 2-complex (or topological space) $X$ a 2-tree, if for every $x,y\in X$, there is a homeomorph of a forest in $X - \{x,y\}$ separating $x$ from $y$. (A forest is a disjoint union of trees.)

Question 1: Has this notion been studied?

Question 2: Is it true that a 2-complex $X$ is a 2-tree iff it contains no homeomorph of a closed surface? (The forward implication is obvious.)

Some examples of 2-trees: standard trees, graphs, $\mathbb{R}^2$, and most interestingly, cartesian products $\mathbb{R} \times T$, where $T$ is any tree (to see this, embed $\mathbb{R} \times T$ in $\mathbb{R}^3$, and cut it with a plane separating $x$ from $y$).

Examples of non-2-trees: closed surfaces, $\mathbb{R}^3$, standard Cayley complex of $\mathbb{Z}^3$ (to see the latter, notice that it contains copies of $\mathbb{S}^2$).

Remark: The definition can be applied recursively, to define $n$-trees for every $n\in \mathbb{N}$.

Update: The answer to Question 2 is negative, as Geva Yashfe points out in the comments.

As Sam Nead points out below, the variant of the definition with “forest” replaced by “tree” seems more sensible. Now a graph is a 2-tree iff it is a tree. Defining a 0-tree to be a point, the recursive definition now starts with 1-trees being the usual trees.

The main question now is: Is every CW-complex (or other nice space, see e.g. Ian Agol’s answer) which is an $n$-tree contractible? Simply-connected?