JavaScript的递归函数调用覆盖

Question

我有一个函数：

function brackets(openStock, closeStock, s) { 
    if (openStock == 0 && closeStock == 0) { 
    document.getElementById('demo').innerHTML = s; 
    } 

    if (openStock > 0) { 
    brackets(openStock - 1, closeStock + 1, s + "("); 
    } 
    if (closeStock > 0) { 
    brackets(openStock, closeStock - 1, s + ")"); 
    } 
} 

应该parenthesizes的组合编写成一段“演示”但是写的第一个之后，其他的人都没有写进段。由于它是一个递归函数，我不知道如何连接结果（组合）。

Answer 1

function brackets(openStock, closeStock,s) 
{ 
    if (openStock == 0 && closeStock == 0) { 
     var e = document.createElement('div'); 
     e.innerHTML = s; 

     while(e.firstChild) { 
      document.getElementById('demo').appendChild(e.firstChild); 
     } 
    } 

    if (openStock > 0) { 
     brackets(openStock-1, closeStock+1, s + "("); 
    } 
    if (closeStock > 0) { 
     brackets(openStock, closeStock-1, s + ")"); 
    } 
}