It is easy to prove that the completion of your space can be any separable metric space where metric spheres are nowhere dense.
Does not it scare you?
Answer to the comment: Not all of your spaces can be embedded into a metric tree.
BTW, there is a nice characterization of subsets in a metric tree: $$ | x - y | + | a - b | \le \max\{|x-a| + |y-b|,|x-b|+|y-a|\}$$ for all points $x,y,a,b$ (here $|x-y|$ denotes the distance from $x$ to $y$). In other words, the values $X=|x-a|+|y-b|$, $Y=|x-b|+|y-a|$ and $Z=|x-y|+|a-b|$ satisfy ulrtatriangle inequality. This inequality can be also thought as a discrete analog of CAT[−∞] inequality.