Data Structures | Tree Traversals | Question 4

What does the following function do for a given binary tree?

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int fun(struct node *root)
{
   if (root == NULL)
      return 0;
   if (root->left == NULL && root->right == NULL)
      return 0;
   return 1 + fun(root->left) + fun(root->right);
}

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(A) Counts leaf nodes
(B) Counts internal nodes
(C) Returns height where height is defined as number of edges on the path from root to deepest node
(D) Return diameter where diameter is number of edges on the longest path between any two nodes.


Answer: (B)

Explanation: The function counts internal nodes.
1) If root is NULL or a leaf node, it returns 0.
2) Otherwise returns, 1 plus count of internal nodes in left subtree, plus count of internal nodes in right subtree.

See the following complete program.

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#include <stdio.h>
  
struct node
{
  int key;
  struct node *left, *right;
};
  
int fun(struct node *root)
{
   if (root == NULL)
      return 0;
   if (root->left == NULL && root->right == NULL)
      return 0;
   return 1 + fun(root->left) + fun(root->right);
}
  
/* Helper function that allocates a new node with the
   given key and NULL left and right pointers. */
struct node* newNode(int key)
{
  struct node* node = (struct node*)
                       malloc(sizeof(struct node));
  node->key = key;
  node->left = NULL;
  node->right = NULL;
  
  return(node);
}
  
/* Driver program to test above functions*/
int main()
{
  
  /* Constructed binary tree is
            1
          /   \
        2      3
      /  \    /
    4     5  8
  */
  struct node *root = newNode(1);
  root->left        = newNode(2);
  root->right       = newNode(3);
  root->left->left  = newNode(4);
  root->left->right = newNode(5);
  root->right->left = newNode(8);
  
  printf("%d", fun(root));
  
  getchar();
  return 0;
}

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