Array flatten function written in ES6 syntax

flatten-array.js

const flattenTco = ([first, ...rest], accumulator) => 
  (first === undefined)
    ? accumulator
    : (Array.isArray(first))
      ? flattenTco([...first, ...rest])
      : flattenTco(rest, accumulator.concat(first))
  
const flatten = (n) => flattenTco(n, []);
  
console.log(flatten([[1,[2,[[3]]]],4,[5,[[[6]]]]]))

var a = [1,2,3,[1,2,3],4,[5,6,7], 9,10] ;
var result = [];
a.forEach(data=>{
	result = [...result].concat(data) 
}) 

console.log(result);

Output
[1, 2, 3, 1, 2, 3, 4, 5, 6, 7, 9, 10]

Not effective but interesting!

arr = [1,2,3,[4,5,[1,2,3,4,56,67],6,7,[7.5,7.6,7.9],8],9,10,11]

const flat = (arr, depth= Infinity ,arr2 = []) =>{
    arr.forEach(e => {
        typeof e === 'object' && depth ? flat(e, depth-1 ,arr2) : arr2.push(e)
    });
    return arr2
}

console.log(flat( arr, 1 )) // [ 1, 2, 3, 4, 5, [ 1, 2, 3, 4, 56, 67 ], 6, 7, [ 7.5, 7.6, 7.9 ], 8, 9, 10, 11 ]  // Depth 1 falt
console.log(flat(arr)) // [ 1, 2, 3, 4, 5, 1, 2, 3, 4, 56, 67, 6, 7, 7.5, 7.6, 7.9, 8, 9, 10, 11 ]  // Full flat

const deepFlat = (arr) => arr.flat(Infinity)

one hack solution. (deep nested also works)
JSON.stringify(arr).split(',').map(each => each.match(/[+-]?\d+(?:\.\d+)?/g)[0])

one hack solution. (deep nested also works)
JSON.stringify(arr).split(',').map(each => each.match(/[+-]?\d+(?:\.\d+)?/g)[0])
work like a charm for number, but not with string ['un', 2, 'trois', 4, 'cinq'].

For immutable solution, this should be the most performant one, because it uses only pop() and push():

const flatten = (list) => { 
  const stack = [list];
  const result = [];

  while(stack.length > 0) {
    const elem = stack.pop();
    if (Array.isArray(elem)) {
      for (let i = elem.length - 1; i >= 0; i--) {
        stack.push(elem[i]);
      }
    } else {
      result.push(elem);
    }
  }

  return result;
};