返回不会退出javascript中的递归函数

Question

I've read the dozen variations on this question, but those answers haven't led me to what must be an obvious mistake. Why does this always return false ? Why do I see called again even after a found it ? And if I put a return in front of the recursive call, why do I never see found it ?

function subResult (object, start, target){
    console.log('called again')
    if (start === target){
      console.log('found it')
      return true
    } else {
      for (var i = 0; i < object[start].edges.length; i++){
        subResult(object, object[start].edges[i], target)
      }
    }
   return false
 }

Answer 1

Change

for (var i = 0; i < object[start].edges.length; i++){
    subResult(object, object[start].edges[i], target)
}

to

for (var i = 0; i < object[start].edges.length; i++){
    if (subResult(object, object[start].edges[i], target)) {
       return true;
    }
}

Ie when found your done. If not keep going.