How to refresh Gulp browser on html change?

Question

css

I want to make a task for Galp that will refresh browser-sync running from the public folder when I change index.html in the source folder .
My gulpfile.js is like this:

gulp.task('build', function() {
    gulp.src('source/index.html')
    .pipe(gulp.dest('public/'))
    .pipe(reload({stream: true}))
})

gulp.task('server', function() {
    browserSync.init({
        server: {
            baseDir: "public/"
        }
    })
})

Well, i.e. copying from source to public works as it should, but changing the source file index.html does not rebuild its copy in the public folder.
How to correctly configure everything for my task?

Answer 1

gulp.task('default', ['server'], function(){

gulp.watch(['путь к файлам html'],  browserSync.reload());

});

Answer 2

Ok, I put it together myself:

gulp.task('buildHTML', function() {
    gulp.src('source/index.html')
    .pipe(gulp.dest('public/'))
    .pipe(reload({stream: true}))
})

gulp.task('watch', function() {
    gulp.watch('source/scss/*.scss', ['compileSass'])
    gulp.watch('source/index.html', ['buildHTML'])
})

gulp.task('server', function() {
    browserSync.init({
        server: {
            baseDir: "public/"
        }
    })
})