CODEWITHSUNDEEP
javascriptrecursion
rswerve
rswerve

Reputation: 179

Return doesn't exit recursive function in javascript

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
 }

Upvotes: 3

Views: 854

Answers (1)

Ed Heal
Ed Heal

Reputation: 60007

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;
    }
}

I.e. when found your done. If not keep going.

Upvotes: 4

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