An idea
Suppose we have a fat Cantor set $\mathcal F\subset[0,1]$, which has empty interior. We can define a function $f$ that is $0$ on $\mathcal F$. The complement of $\mathcal F$ in $[0,1]$ is a union of open intervals. For each interval, let $f$ be a continuous function that looks like "$\land$" on it with $0$ at boundaries.
This idea is also stated by as Sassatelli Giulio.
A fat cantor set
For any $a=\sum _{k=1}^\infty a_{k}2^{-k^2}\in[0,1]$, where $a_k=\lfloor2^{k^2}a-2^{2k-1}\lfloor2^{(k-1)^2}a\rfloor\rfloor\in\{0,1, 2, \cdots, 2^{2k-1}-1\}$. In terms of positional notation system,
$$a=a_1a_2a_3a_4\cdots,$$
where the weight of each position is respectively $\frac1{2^1},\frac1{2^4},\frac1{2^9},\frac1{2^{16}},\cdots.$
Consider $\mathcal F=\{\sum _{k=1}^\infty a_{k}2^{-k^2}: a_k\in\{0,1, 2, \cdots, 2^{2k-1}-1\}, a_k\not=2^{2k-2}\}\subset[0,1]$. In plain words, we start with $[0,1]$. At round $k\in\{1,2,\cdots\}$, for each remaining intervals, we split it into $2^{2k-1}$ equal pieces and then take away the interior of one of the pieces in the middle. All the points that have never been taken away form $\mathcal F$.
$\ \mathcal F$ is a fat Cantor set with Lebesgue measure
$$\prod_{k=1}^{\infty}\frac{2^{2k-1}-1}{2^{2k-1}}>1-\sum_{k=1}^{\infty}\frac1{2^{2k-1}}=\frac13.$$
A continuous function $f$ with $f^{-1}(0)$ having positive lebsegue measure
Suppose $a\in[0,1], a\notin\mathcal F$. Since $\mathcal F$ is a closed subset of $\Bbb R$, there are open intervals that are disjoint with $\mathcal F$ that contain $a$. Let $(\ell_a,r_a)$ be the biggest one of them.
Define $f:[0,1]\to\Bbb R$,
$$f(a)=\begin{cases}
0&\text{if }a\in\mathcal F,\\
(\ell_a+r_a)/2-|a-(\ell_a+r_a)/2|&\text{if }a\notin\mathcal F.\\
\end{cases}$$
$f$ is continuous.
$f^{-1}(a)$ is finite if $a\not=0$.
$f^{-1}(0)=\mathcal F$, a fat Cantor set with Lebseque measure $>\frac13$.
All level sets of $f$ have empty interiors.
