Generate all balanced bracket combinations

Print all possible balanced parenthesis combinations up to N. For example:

N = 2
(()), ()()

N = 3
((())), (()()), (())(), ()(()), ()()()

We will implement a recursive function to solve this challenge. The idea is: 1. Add a left bracket to a newly created string. 2. If a left bracket was added, potentially add a new left bracket and add a right bracket. 3. After each of these steps we add the string to an array that stores all bracket combinations.

var all = [];

function parens(left, right, str) {
  
  // if no more brackets can be added then add the final balanced string
  if (left === 0 && right === 0) {
    all.push(str);
  }
  
  // if we have a left bracket left we add it
  if (left > 0) {
    parens(left-1, right+1, str+"(");
  }
  
  // if we have a right bracket left we add it
  if (right > 0) {
    parens(left, right-1, str+")"); 
  }
  
}

// the parameters parens(x, y, z) specify:
// x: left brackets to start adding
// y: right brackets we can add only after adding a left bracket
// z: the string so far
parens(3, 0, "");
console.log(all);