Here are my thoughts: If you write $c_0 \equiv |c_0|\, e^{i\phi_0}$ and $c_1 \equiv |c_1|\, e^{i\phi_1}$, then you can write the wave function as

$$|\Psi\rangle = e^{i\phi_0}\left(|c_0| \,|0\rangle + |c_1|\,e^{i(\phi_1-\phi_0)}\,|1\rangle\right)\quad . $$

The associated density operator is then given by $\rho_{\Psi}\equiv |\Psi\rangle\langle \Psi|$. The diagonal elements will yield $|c_0|$ and $|c_1|$ and from the off-diagonal terms you can reconstruct $|c_1| \, e^{i(\phi_1-\phi_0)}$. However, you can only reconstruct the wave function up to the global phase, which is also intuitive, since two wave functions $|\Psi\rangle$ and $|\psi\rangle$ which differ only by a global phase will yield the same density operator.

If you write $c_0 \equiv |c_0|\, e^{i\phi_0}$ and $c_1 \equiv |c_1|\, e^{i\phi_1}$, then you can write the wave function as

$$|\Psi\rangle = e^{i\phi_0}\left(|c_0| \,|0\rangle + |c_1|\,e^{i(\phi_1-\phi_0)}\,|1\rangle\right)\quad . $$

The associated density operator is then given by $\rho_{\Psi}\equiv |\Psi\rangle\langle \Psi|$. The diagonal elements will yield $|c_0|$ and $|c_1|$ and from the off-diagonal terms you can reconstruct $|c_1| \, e^{i(\phi_1-\phi_0)}$. However, you can only reconstruct the wave function up to the global phase, which is also intuitive, since two wave functions $|\Psi\rangle$ and $|\psi\rangle$ which differ only by a global phase will yield the same density operator.