Of course it seems you meant topological genus, but here is a similar argument for the arithmetic genus, that I learned from a paper of Fulton. Maybe this is related to the adjunction argument you refer to. And if you combine this with Francesco's answer, you get a proof that the two are equal for smooth plane curves, (or if you assume that result, the Hirzebruch-Riemann-Roch theorem for curves, then you get an answer to your question).

**Lemma A:** If $X,Y$ are two curves on a smooth surface $S$, and if $X,Y$ are linearly equivalent as divisors on $S$, then $\chi(\mathcal{O}_X) = \chi(\mathcal{O}_Y)$.

*Remark*: This says in some sense $\chi(\mathcal{O})$ is a deformation invariant, at least for linear deformations.

**Proof:** Since the line bundles $\mathcal{O}_S(-X)$ and $\mathcal{O}_S(-Y)$ are isomorphic on $S$, the invariants $\chi(\mathcal{O}_S(-X))$ and $\chi(\mathcal{O}_S(-Y))$ are equal. By the usual exact sheaf sequence $$0\to \mathcal{O}_S(-X)\to \mathcal{O}_S\to \mathcal{O}_X\to 0$$ and the analogous one for $Y$, plus the additivity of $\chi$, we get that
\begin{align}
\chi(\mathcal{O}_X)
&= \chi(\mathcal{O}_S) - \chi(\mathcal{O}_S(-X)) \\
&= \chi(\mathcal{O}_S) - \chi(\mathcal{O}_S(-Y)) \\
&= \chi(\mathcal{O}_Y).\hspace{1cm}&\text{qed.}\end{align}

**Lemma B:** Now suppose that $Y, Y'$ are curves on a smooth surface $S$, and that $Y$ and $Y'$ meet transversely at precisely $n$ points. Then we claim $\chi(\mathcal{O}_{Y+Y'}) = \chi(\mathcal{O}_Y) + \chi(\mathcal{O}_{Y'}) - n.$

**Proof:** Consider the sequence
$$0\to \mathcal{O}_{Y+Y'}\to \mathcal{O}_Y + \mathcal{O}_{Y'}\to O_{Y\cdot Y'}\to 0,$$ induced by the map from the disjoint union of $Y,Y'$, to their union $Y+Y'$ on $S$, and where the map to $\mathcal{O}_{Y\cdot Y'}$ is the difference of the two restrictions, from $Y$ and from $Y'$, to the intersection of $Y$ and $Y'$. The additivity of $\chi$ then implies the desired relation, i.e.
\begin{align}
\chi(\mathcal{O}_Y) + \chi(\mathcal{O}_Y')
&= \chi(\mathcal{O}_Y + \mathcal{O}_{Y'}) \\
&= \chi(\mathcal{O}_{Y+Y'}) + \chi(\mathcal{O}_{Y\cdot Y'}) \\
&= \chi(\mathcal{O}_{Y+Y'}) + n.
\end{align} Thus $\chi(\mathcal{O}_{Y+Y'}) = \chi(\mathcal{O}_Y) + \chi(\mathcal{O}_{Y'}) - n.$ qed.

Now that we know how the function chi(O) behaves under linear degeneration, all we need is to find a formula that behaves this way, and it must be the formula for $chi(\mathcal{O})$.

**Corollary:** If $X$ is a smooth plane curve of degree $d$, then
$\chi(\mathcal{O}_X) = 1 - \frac12 (d-1)(d-2).$

**Proof:** Induction on $d$. If $d = 2$, then the smooth conic $X$ moves in a linear series also containing a union $Y$ of two lines $Y_1 + Y_2$, where each line is isomorphic to $X$. Then by lemmas A,B above, we have
$$\chi(X) = \chi(Y_1)+\chi(Y_2) - 1 = \chi(X)+\chi(X)-1,$$ hence $\chi(X) = 1$. This proves the case $d = 2$, and since a smooth curve of degree $d = 1$ is isomorphic to one of degree 2, we also obtain the formula for degree $d=1$.

Now assume $d \geq 3$ and that we have proved the formula for smooth curves
of degree $<d$. A smooth degree-$d$ curve $X$ moves in a linear series that also
contains a curve of form $Y = Y_1+Y_2,$ where $Y_1$ is smooth of degree $d-1$, and $Y_2$ is a line meeting $Y_1$ transversely in $d-1$ distinct points. Then lemmas A, B and induction give us that
\begin{align}
\chi(\mathcal{O}_X)
&= \chi(\mathcal{O}_Y) \\
&= \chi(\mathcal{O}_{Y_1})+\chi(O_{Y_2})-(d-1) \\
&= 1-\frac12 (d-2)(d-3) + 1 - (d-1) \\
&= 1-\frac12 (d-1)(d-2),\end{align} as desired. qed.