return current node from BST

Question

Hello I've run into an issues I cannot seem to resolve. I have a BST that I am traversing through and checking ranks. I have a method checkRank(link head, targRank) that takes in the head node and traverses through the tree until it finds a node with an equal rank to targRank. What I am trying to do is have the checkRank function return the current node it found the equal rank at. What would be the best way to achieve this because all my attempts seem to return the current node as the head?

typedef struct node* link;

struct node 
{
    Item item;  // Data for this node
    link l, r;  // left & right links
    int rank;
};

Func call:

link head;
checkRank(head, 13);

Func:

link checkRank(link h,int targetRank)
{
    if (h != NULL)
    {
        if (h->rank < targRank)
        {
            checkRank(h->r, targRank);
        }


    if (h->rank > tarRank)
        {
            checkRank(h->l, targtRank);
        }

        if (h->rank == targRank)
        {
            return ??;
        }
    }
    else
    {
        printf("Equiv rank could not be found\n");
    }
}

score 1 · Accepted Answer · answered Apr 18 '13 at 02:50

First of all, you need to return something along each path. Have you considered something like the following:

link check_rank(link h, int target) {
  if (h == NULL) {
    printf("equivalent rank could not be found\n");
    return NULL;
  }
  if (h->rank < target)
    return check_rank(h->r, target);
  if (h->rank > target)
    return check_rank(h->l, target);
  return h;
}

Functions have to always return a value and many recursive functions will follow the pattern of (1) return a sentinel to stop recursion when the appropriate condition is met or (2) recurse and return whatever the recursive call returns.

return current node from BST

1 Answers1