What is the most general matrix that gives rise to all even characteristic polynomials?

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Is there some general form of all matrices which give rise to all even or all odd characteristic polynomial terms?

For skew-symmetric matrices such that $A^T=-A$ we necessarily have all even or all odd terms.

For example the matrix $$A=\left(\matrix{0&2&4\\-2&0&-1\\-4&1&0}\right)$$ has the characteristic polynomial $$-\lambda^3-21\lambda$$

However we can also generate characteristic polynomials of this form when the matrix is not skew symmetric.

It appears that we require the trace to be $0$, so that the sum of eigenvalues are equal to $0$. But this doesn't appear to be a general condition.

An example would be the matrix $$B=\left(\matrix{3&0&2&4&0&0\\0&-3&0&5&-2&-4\\2&0&3&4&0&-5\\4&5&4&7&0&0\\0&-2&0&0&-3&-4\\0&-4&-5&0&-4&-7}\right)$$

which has the characteristic polynomial

$$\lambda^6-189\lambda^4+2472\lambda^2-6084$$