-、1. D.
From the problem, we know that $A=[0,2], B=\{0,1\}$. Therefore, $A \cap B=\{0,1\}$.
$$
\begin{array}{l}
A=\left\{x \mid x^{2}-2 x \leqslant 0\right\}, \\
B=\left\{x \left\lvert\, \frac{x+1}{x-2} \leqslant 0\right., x \in \mathbf{N}\right\} .
\end{array}
$$

Then $A \cap B=(\quad)$.
(A) $[0,2]$
(B) $[0,2)$
(C) $\{1\}$
(D) $\{0,1\}$