Counting the number of leaf nodes in a binary tree

http://www.geeksforgeeks.org/iterative-program-count-leaf-nodes-binary-tree/

unsigned int getLeafCount(struct Node* node)

{

    // If tree is empty

    if (!node)

        return 0;

 

    // Initialize empty queue.

    queue<Node *> q;

 

    // Do level order traversal starting from root

    int count = 0; // Initialize count of leaves

    q.push(node);

    while (!q.empty())

    {

        struct Node *temp = q.front();

        q.pop();

 

        if (temp->left != NULL)

            q.push(temp->left);

        if (temp->right != NULL)

            q.push(temp->right);

        if (temp->left == NULL && temp->right == NULL)

            count++;

    }

    return count;

}

Problem solving with programming: Counting the number of leaf nodes in a binary tree
Given a binary tree, write a program to count the number of leaf nodes. A leaf node is a node which does not have left and right children.
http://www.geeksforgeeks.org/write-a-c-program-to-get-count-of-leaf-nodes-in-a-binary-tree/

unsigned int getLeafCount(struct node* node)

{

  if(node == NULL)      

    return 0;

  if(node->left == NULL && node->right==NULL)     

    return 1;           

  else

    return getLeafCount(node->left)+

           getLeafCount(node->right);     

}

Use Queue ===>we can use Queue. Idea is to use Level Order traversal.
http://k2code.blogspot.com/2015/09/program-to-count-leaf-nodes-in-binary.html

int countLeaves(Node root)

{

  int count=0;

    if(root==null)

      return 0;

    Queue<Node> myqueue = new Queue<Node>();

    myqueue.push(root);

    while(!myqueue.empty())

    {

      Node temp;

       temp=myqueue.pop();//take the top element from the queue

      if(temp.left==null && temp.right==null)

       count++;

      if(temp.left!=null)

       myqueue.push(temp.left);

      if(temp.right!=null)

       myqueue.push(temp.right);

    }

  return count;

}

http://comproguide.blogspot.com/2014/05/counting-number-of-leaf-nodes-in-binary.html

int leafNodesIterative(TreeNode *root )


{


if( root == NULL )


return 0;




int count = 0;


queue<TreeNode*> Q;


Q.push( root );




while ( !Q.empty() )


{


TreeNode * temp = Q.front();


Q.pop();




if( temp->left == NULL && temp->right == NULL )


count++;




if( temp->left !=  NULL )


Q.push( temp->left );


 


if( temp->right !=  NULL )


Q.push( temp->right );


}


 


return count;


}

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