Question

Doing a project with BSTs, I have a logic fault somewhere in my insertion function, just can’t seem to find it.

Insertion function:

int bst_insert (bst_t *tree, bst_key_t key)
{
  bst_node_t *node, *temp_node;
  if( tree == NULL ) {
    printf("Invalid tree pointer.\n");
    return;
  }
  if( tree->root == NULL ) {
    node = (bst_node_t *)malloc(sizeof(bst_node_t));
    node->left = node->right = NULL;
    node->key = key;
    tree->root = node;
    return 1;
  }
  temp_node = tree->root;
  while(1) {
    if( temp_node == NULL ) {
      node = (bst_node_t *)malloc(sizeof(bst_node_t));
      node->left = node->right = NULL;
      node->key = key;
      temp_node = node;
      return 1;
    }
    if( temp_node->key == key ) {
      temp_node->data_ptr = data;
      return 0;
    } else if( key < temp_node->key ) {
      temp_node = temp_node->left;
    } else if( temp_node->key < key ) {
      temp_node = temp_node->right;
    }
  }  
}

In using this function, inserting one node works just fine (probably because the tree->root is null, so it exits inside that if statement), but when I try and insert a second and leave this function, the tree only has the first node in it.

Let me know if I forgot to give any pertinent information.

Answer 1

The problem is here:

temp_node = tree->root;
...
temp_node = node;

You’re assuming that when you perform the second assignment, it actually affects the node temp_node was pointing to. But in fact, it simply replaces the contents of temp_node without changing anything else (it’s a local variable).

One solution is to have a pointer to a “node pointer”. Then, when you change the value that this pointer is pointing to, you’re actually “changing the world” instead of changing a local variable. Something like:

bst_node_t **temp_node;
...
temp_node = &tree->root;

/* To change the value that temp_node is pointing to */
*temp_node = node;

/* To change temp_node itself */
...
} else if( key < temp_node->key ) {
  temp_node = &temp_node->left;
} else if( temp_node->key < key ) {
  temp_node = &temp_node->right;
}

It is possible to fix this without using a pointer-to-pointer, but then you have to keep remembering the previous (parent) pointer, and whether you need to change left or right.