http://www.geeksforgeeks.org/iterative-program-count-leaf-nodes-binary-tree/
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/
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
http://comproguide.blogspot.com/2014/05/counting-number-of-leaf-nodes-in-binary.html
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;}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); }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;}Read full article from Problem solving with programming: Counting the number of leaf nodes in a binary treeint 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;}
