In my textbook (NCERT 12th class Chemistry, Part 2, India), $\ce{H2}$ with $\ce{Pd}$ catalyst is the only method mentioned.
3 Answers
Yes, $\ce{LiAlH4}$ can be used for the reduction of an aldehyde to an alcohol. Compared to $\ce{NaBH4}$ it is a stronger reducing agent due to the electronegativity differences (Pauling electronegativity of $\ce{Al}$ is 1.61 & $\ce{B}$ is 2.04) making the metal-H bond more polar in case of $\ce{LiAlH4}$. Therefore, it reacts very violently with water and other acidic group containing compounds, and careful addition of $\ce{LiAlH4}$ is required.
The mechanism is rather straightforward, with alkoxide formation and protonation steps to get the final product.
In the lithium aluminum hydride reduction, water is usually added in a second step, whereas in the sodium borohydride reduction the methanol solvent system achieves this protonation immediately.
It really depends on the molecule itself. If you're talking about a molecule where the aldehyde is virtually the only functional group present, then yes $\ce{LiAlH4}$ will be very efficient. However, in the presence of other sensitive functional groups such as nitriles which can also be reduced with $\ce{LiAlH4}$, using milder reagents such as $\ce{NaBH4}$ makes more sense since nitriles cannot be reduced by it.
I would suggest you to kindly refer page no. 360 of Chemistry Part II NCERT of class 12.
The paragraph having title Reduction reads:
Reduction to alcohols: Aldehydes and ketones are reduced to primary and secondary alcohols respectively by sodium borohydride $\ce{NaBH4}$ or lithium aluminium hydride $\ce{LiAlH4}$ as well as by catalytic hydrogenation.