What is length of the universe on lights perspective According to Einsteins theory of special relativity, when something is travelling close to light speed, travel path is contracted. When the traveller is light itself, does it see the travel path contracted to zero? In another word, does light see length of the whole universe is zero across its travel path? How?
Sort of.  The Lorentz factor is

$$
\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}
$$

whereupon a stationary object in the stationary reference frame of length $L$ has a length of $L' = \frac{L}{\gamma} = L\sqrt{1-\frac{v^2}{c^2}}$ in the moving reference frame.  As the velocity $v$ increases toward $c$, we get

$$
\lim_{v \to c} L' = L \lim_{v \to c} \sqrt{1-\frac{v^2}{c^2}} = 0
$$

That is to say, if a spaceship were to move very rapidly from here to the center of the galaxy and back again, a clock aboard that spaceship would advance much less than an initially synchronized clock here at home.  From our point of view, the reason is that the clock on board the spaceship slowed down (time dilation).  From the point of view of someone on the spaceship, the reason is that the distance to the center of the galaxy was shortened (length contraction).  And in the limit (which cannot actually be reached by anything with mass), the clock on the spaceship becomes arbitrarily slow from our point of view, while the distance to the center of the galaxy becomes arbitrarily short from their point of view.  Of course, we can't actually ask the photon how it experiences the trip (even in principle), since the trip is instantaneous for it.  So the way to express it is to say what the limiting behavior is, as $v$ approaches $c$.

Wikipedia has a useful article on special relativity.