1. We are given that $\alpha, \beta$ are the distinct roots of $x^2-x+1=0.$

2. Solve the quadratic: $x=\frac{1\pm \sqrt{1-4}}{2}=\frac{1\pm i\sqrt{3}}{2}.$ So the roots are $\alpha=\frac{1+i\sqrt{3}}{2},\qquad \beta=\frac{1-i\sqrt{3}}{2}.$ These are complex cube roots of unity other than $1$.

3. From $x^2-x+1=0,$ for a root $x$, we get $x^2-x+1=0 \implies x^2=x-1.$ Multiplying by $(x+1)$, $x^3+1=0 \implies x^3=-1.$ Hence $x^6=1.$ Therefore, $\alpha^6=1,\qquad \beta^6=1.$

4. Reduce the exponents modulo $6$: $101 \equiv 5 \pmod{6}, \qquad 107 \equiv 5 \pmod{6}.$ So, $\alpha^{101}=\alpha^5,\qquad \beta^{107}=\beta^5.$ Thus $\alpha^{101}+\beta^{107}=\alpha^5+\beta^5.$

5. Since $\alpha^6=1$ and $\beta^6=1$, $\alpha^5=\alpha^{-1},\qquad \beta^5=\beta^{-1}.$ Therefore, $\alpha^5+\beta^5=\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}.$

6. Using Vieta's formulas for $x^2-x+1=0,$ we have $\alpha+\beta=1,\qquad \alpha\beta=1.$ Hence, $\frac{\alpha+\beta}{\alpha\beta}=\frac{1}{1}=1.$

7. Therefore, $\alpha^{101}+\beta^{107}=1.$

So the correct option is D.