How to set gulp.dest() in same directory as pipe inputs?

I need all the found images in each of the directories to be optimized and recorded into them without setting the path to the each folder separately. I don't understand how to make that.

var gulp = require('gulp');
var imageminJpegtran = require('imagemin-jpegtran');

gulp.task('optimizeJpg', function () {

return gulp.src('./images/**/**/*.jpg')
    .pipe(imageminJpegtran({ progressive: true })())
    .pipe(gulp.dest('./'));
});

Here are two answers.
First: It is longer, less flexible and needs additional modules, but it works 20% faster and gives you logs for every folder.

var merge = require('merge-stream');

var folders =
[
    "./pictures/news/",
    "./pictures/product/original/",
    "./pictures/product/big/",
    "./pictures/product/middle/",
    "./pictures/product/xsmall/",
    ...
];

gulp.task('optimizeImgs', function () {

    var tasks = folders.map(function (element) {

        return gulp.src(element + '*')
            .pipe(sometingToDo())
            .pipe(gulp.dest(element));

    });

    return merge(tasks);

});

Second solution: It's flexible and elegant, but slower. I prefer it.

return gulp.src('./pictures/**/*')
    .pipe(somethingToDo())
    .pipe(gulp.dest(function (file) {
        return file.base;
    }));

Here you go:

gulp.task('optimizeJpg', function () {

    return gulp.src('./images/**/**/*.jpg')
        .pipe(imageminJpegtran({ progressive: true })())
        .pipe(gulp.dest('./images/'));
});

Gulp takes everything that's a wildcard or a globstar into its virtual file name. So all the parts you know you want to select (like ./images/) have to be in the destination directory.

You can use the base parameter:

gulp.task('uglify', function () {
  return gulp.src('./dist/**/*.js', { base: "." })
    .pipe(uglify())
    .pipe(gulp.dest('./'));
});