2. Let $x, y, z$ be non-negative real numbers, and $x+y+z=$ 2. Then the sum of the maximum and minimum values of $x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}$ is $\qquad$ .
2.1.

Since $x, y, z$ are non-negative real numbers, we have
$$
A=x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2} \geqslant 0.
$$

When $x=y=0, z=2$, $A=0$, thus the minimum value of $A$ is 0.

Assume $A$ reaches its maximum value at $(x, y, z)$, without loss of generality, let $x \leqslant y \leqslant z$, prove: $x=0$.

In fact, if $x>0$, then let $x^{\prime}=0, y^{\prime}=y+x, z^{\prime}=z$, at this time,
$$
\begin{array}{l}
A^{\prime}=x^{\prime 2} y^{\prime 2}+y^{\prime 2} z^{\prime 2}+z^{\prime 2} x^{\prime 2}=(y+x)^{2} z^{2} \\
=y^{2} z^{2}+z^{2} x^{2}+2 x y z^{2} \\
\geqslant y^{2} z^{2}+z^{2} x^{2}+2 x y(x y) \\
>y^{2} z^{2}+z^{2} x^{2}+x y(x y)=A,
\end{array}
$$

which contradicts the maximality of $A$.
When $x=0$,
$$
y+z=2, A=y^{2} z^{2} \leqslant\left(\frac{y+z}{2}\right)^{4}=1,
$$

the equality holds when $y=z=1, x=0$.
Therefore, the maximum value of $A$ is 1.
Thus, the sum of the maximum and minimum values is 1.