It is a well-known theorem first proved by Bourgain that any map $\varphi:T\_n\to H$ from the binary tree of height $n$ to a Hilbert space has distortion at least $C \sqrt{\ln n}$ where $C$ is a universal constant. Distortion is the least $D$ such that for some $s$ and all $x,y\in T\_n$, we have
$$ s d(x,y) \leq d(\varphi(x),\varphi(y)) \leq sD d(x,y)$$
(This also generalizes in some way to many more Banach spaces).


Some of the proofs use the linear structure of $H$, but others are purely metric. I therefore wonder:



> 
> Has this distortion estimate been generalized to spaces of non-negative curvature (Alexandrov spaces)?
> 
> 
> 



Distortion of tree embedding in Alexandrov spaces
See the direct proof of markov convexity in proposition 2.1 of the following paper


Mendel, Manor; Naor, Assaf Markov convexity and local rigidity of distorted metrics. J. Eur. Math. Soc. (JEMS) 15 (2013), no. 1, 287–337. 


the only thing that is used there is lemma 2.3 of the above references, which holds with p=2 for alexandrov spaces of nonnegative curvature. Therefore the answer to your question is known to be positive. The observation that the proof of the above reference works in this setting has been shown in talks that I heard over the past few years, and also stated explicitly in the following paper


Sean Li, Coarse differentiation and quantitative nonembeddability for Carnot groups 


