I tried solving using recursion but am not sure whether my answer is correct.
I tried using P(N) = P(N-1)*0.25*1 (For the last two TT) + P(N-1)*0.25*1 (For the last two HT) + P(N-1)*0.25*1 (For the last two TH) + P(N-1)*0.25*0.5 (For the last two HH, since it can be followed only by T)
Let's count the toss sequences with no triple $HHH$. Let $T_n$ be the number of such, of length $n$. Of course, the answer you seek is $\frac {T_n}{2^n}$.
Let $A_n$ be those "good" sequences of length $n$ that end in $T$, let $B_n$ be those that end in $TH$, and let $C_n$ be those that end in $THH$. Then $$T_n=A_n+B_n+C_n$$ Furthermore, $$A_n=T_{n-1}\quad B_n=A_{n-1}=T_{n-2}\quad C_n=B_{n-1}=T_{n-2}$$ Whence $$T_n=T_{n-1}+T_{n-2}+T_{n-3}$$ Clearly we have $T_1=2,\;T_2=4,\;T_3=7$
I'm not aware of a pleasant closed formula, though of course standard methods will give an expression in terms of the roots of the associated cubic. See this for some details. Of course it's easy to implement the recursion for modest $N$.