Finding a vertex of a tree such that every subtree rooted at that vertex is of a uniform color
Some comments on the code:

for i in range(n-1):
    (u, v) = [int(i) for i in input().split()]


You are using i as the outer loop index, and i as the inner (list comprehension) loop index.  This is confusing, and should be discouraged.

The second line may be written more simply as:

    u, v = map(int, input().split())


The assignment:

colors = defaultdict(lambda: -1)


is overwritten by the code that creates the colors dictionary (below), and so can be removed.

You are changing the variable meaning in the following lines:

c = [int(i) for i in input().strip().split()]
colors = {i+1: v for (i, v) in enumerate(c)}


In the first line, i is a colour value; in the second line, v is the colour value and i is the index of the colour value.  For consistency, I'd used v instead of i in the first line.  Or get rid of the list comprehension index altogether:

c = map(int, input().strip().split())
colors = { i: v for i, v in enumerate(c, 1) }


Using enumerate(c, 1) eliminates the need to use i+1 as the key expression.

Is the .strip() necessary?



I'm not certain what the point of maintaining your visited set is.  Each time you start() a search from a new point, you are resetting the the visited set, and during any dfs(), you won't encounter a node you've already visited because the graph is guaranteed to be be a tree.  You just need to pass the start node for each dfs() step as a previous node (instead of a visited set) to prevent back-tracking. 



Use better variable names.   I just figured out the nebs is an abbreviation for neighbours. 



Faster solution hint:

I don’t believe a DFS is necessary.


Ignore all edges between vertices of the same colour
All other edges must share a common vertex (or Timofey will be annoyed)
Examine first two edges between different coloured vertices.  Continue with a linear search of remaining edges