# Counting Nodes in a Binary Tree

### Count All Nodes in Binary Tree :

int countAllNodes(struct Node *ptr) {
// Variables to obtain nodes is left and right subtree
int leftNodes, rightNodes;
if (ptr) {
leftNodes = countAllNodes(ptr -> left);
rightNodes = countAllNodes(ptr -> right);
return leftNodes + rightNodes + 1;
}
return 0;
}

### Count Nodes Having Both Children (Degree 2) :

int countTwoChildren(struct Node *ptr) {
// Variables to obtain nodes is left and right subtree
int leftNodes, rightNodes;
if (ptr) {
leftNodes = countTwoChildren(ptr -> left);
rightNodes = countTwoChildren(ptr -> right);
// Check if both children are present
if (ptr -> left && ptr -> right) {
return leftNodes + rightNodes + 1;
} else {
return leftNodes + rightNodes;
}
}
return 0;
}

### Count Nodes Having One or More Children (Degree 1 or 2) :

int countOneOrMore(struct Node *ptr) {
// Variables to obtain nodes is left and right subtree
int leftNodes, rightNodes;
if (ptr) {
leftNodes = countOneOrMore(ptr -> left);
rightNodes = countOneOrMore(ptr -> right);
// Check if atleast one child is present
if (ptr -> left || ptr -> right) {
return leftNodes + rightNodes + 1;
} else {
return leftNodes + rightNodes;
}
}
return 0;
}

### Count Nodes Having Exactly One Child (Degree 1) :

int countOneChild(struct Node *ptr) {
// Variables to obtain nodes is left and right subtree
int leftNodes, rightNodes;
if (ptr) {
leftNodes = countOneChild(ptr -> left);
rightNodes = countOneChild(ptr -> right);
// Perform XOR operation to check if only child is present
if (ptr -> left ^ ptr -> right) {
return leftNodes + rightNodes + 1;
} else {
return leftNodes + rightNodes;
}
}
return 0;
}

### Count Leaf Nodes (Degree 0) :

int countLeafNodes(struct Node *ptr) {
// Variables to obtain nodes is left and right subtree
int leftNodes, rightNodes;
if (ptr) {
leftNodes = countLeafNodes(ptr -> left);
rightNodes = countLeafNodes(ptr -> right);
// Check if both children are NULL
if (!(ptr -> left) && !(ptr -> right)) {
return leftNodes + rightNodes + 1;
} else {
return leftNodes + rightNodes;
}
}
return 0;
}

### Alternative Way of Counting (Without Variables) :

int count(struct Node *ptr) {
if (ptr == NULL) {
return 0;
}
return count(ptr -> left) + count(ptr -> right) + 1;
}
Following the same style for all the other conditions.
Contributed by Nitin Ranganath