二叉树，返回节点的父级

Question

I try to write a function which return the parent of the value node.The createAB function works,but i don't know to iterate the binary tree elements.How to make recursively call?.Please help me.

 ifstream f("store.txt");//store.txt:10 7 8 0 0 0 13 9 0 11 0 0 12 0 0
    struct elem {
        int inf;
        elem* st;
        elem* dr;
    };
    //this function create the binary tree
    void createAB(elem*& p) {
        int n;
        f >> n;
        if (n!=0) {
            p = new elem;
            p->inf = n;
            createAB(p->st);
            createAB(p->dr);
        }
        else 
            p = NULL;
    }
    `

        elem* parent(elem* rad, int n) {//my function,doesn't work
        if (rad == NULL)
            return NULL;
        else
            if (rad->st->inf == n || rad->dr->inf == n) 
                return rad;//return the element
            else {
                return parent(rad->st, n);//here is a problem
                return parent(rad->dr, n);
            }
    }
      10
   7       13
 8       9    12
           11
node 12 => parent 13
node 8 => parent 7

Answer 1

在使用rad->st->inf和rad->dr->inf之前，应检查rad->st和rad->dr是否为null。

Answer 2

As pointed out by sebi519, you need to check that rad->st and rad->dr are not NULL .

But as you correctly commented the main problem here is in the return statements. The problem is that the second line is never executed as the first line already returns from the function.

return parent(rad->st, n);//This returns and the next line is not executed
return parent(rad->dr, n);

You need to save the result of the first call and check if it was successful and if not then do the second call:

elem* result = parent(rad->st,n);
if (result != NULL)
  return result;
else
  return parent(rad->dr,n);

Full corrected function:

elem* parent(elem* rad, int n) {
  if (rad == NULL)
    return NULL;
  else
    if ( (rad->st!=NULL && rad->st->inf == n) || (rad->dr!=NULL) && (rad->dr->inf == n)){
      return rad;//return the element
    }
    else {
      elem* result= parent(rad->st, n);
      if (result!=NULL){
        return result;
      }
      else{
        return parent(rad->dr, n);
      }
    }
}