Is the below language Non R.E? $L_0=\{\langle M,w,0\rangle\mid M \text{ halts on } w\}$

$L_1=\{⟨M,w,1⟩\mid M \text{ does not halts on } w\}$

Here $\langle M,w,i \rangle$ is a triplet, whose first component $M$ is an encoding of a Turing Machine, second component $w$ is a string, and third component $i$ is a bit. 

Let $L=L_0 \cup L_1$. Is $L$ non R.E ?



After seeing the question I was able to figure out that there are some strings that do not belong to $L_0$ as well as $L_1$.

Like, lets take the case for $L_0$ and a string $001\dots10−01−1$, ("$-$" shown for notation purpose only) where the  first component describes a TM $M$ followed by input "$w=01$" and last bit "$1$". Now suppose M halts on "$01$". Still the given input is not in $L_0$ as the last bit is "$1$" and not "$0$" as required by $L_0$. So, this input must be in $L_1$. But since $M$ halts on $w$, this input is not in $L_1$ either.  So there are infinite strings like these. I am not able to prove that these infinite  set of strings are not r.e.

Is there any method to solve these kind of problems? I am not able to master these kind of problems.

Please someone help.
Assume that $L$ is r.e. and $M_L$ recognizes $L$. Given $\langle M,w \rangle$, $M$ on $w$ either halts or not. This means that either $\langle M, w, 0 \rangle$ or $\langle M, w, 1 \rangle$ must be in $L$. Then using a universal Turing machine start to run/simulate $M_L$ on both inputs $\langle M, w, 0 \rangle$ and $\langle M, w, 1 \rangle$, simultaneously, say one step at a time for each input. After each step check if $M_L$ halts on at least one input. Since $L$ is r.e. and exactly one of $\langle M, w, 0 \rangle$ and $\langle M, w, 1 \rangle$ belongs to $L$, $M_L$ must eventually halt on one of them. Depending on which input $M_L$ halts, you can tell if $M$ halts on $w$. This decides the Halting problem and hence $L$ is not r.e.