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Data Structures
  • Data Structures Manual
  • Arrays
    • Array ADT
    • Linear Search
    • Binary Search
    • Some More Basic Operations
    • Reversing an Array
    • Operations in a Sorted Array
    • Merging Two Arrays
    • Set Operations
    • Finding Missing Elements
    • Duplicates in an Array
    • Getting a Pair whose Sum = K
    • Finding Max & Min in Single Scan
  • Strings
    • Finding the Length of a String
    • Changing Cases in a String
    • Finding Number of Vowels, Consonants & Words
    • Reversing a String
    • Checking for Palindrome
    • Duplicates in a String
    • Checking if Strings are Anagrams
    • Permutations of a String
  • Singly Linked List
    • Displaying the Nodes
    • Counting the Nodes
    • Sum of all Nodes
    • Finding the Maximum Element
    • Searching in a Node
    • Inserting a Node
    • Inserting a Node in Sorted List
    • Deleting a Node
    • Checking if List is Sorted
    • Removing Duplicates from a List
    • Reversing a Linked List
    • Concatenating Two Lists
    • Detecting a Loop in Linked List
    • Merge Two Sorted Lists
    • Finding the Middle Node
  • Cirular Linked List
    • Displaying the Nodes
    • Inserting a Node
    • Deleting a Node
  • Doubly Linked List
    • Inserting a Node
    • Deleting a Node
    • Reversing a Doubly Linked List
    • Circular Doubly Linked List
  • Stack
    • Stack Using Array
    • Stack Using Linked List
    • Balancing Parenthesis
    • Infix to Postfix
    • Evaluation of Postfix Expression
  • Queue
    • Queue using Array
    • Queue using Linked List
    • Double Ended Queue
  • Binary Tree
    • Creating a Binary Tree using Queue
    • Recursive Tree Traversals
    • Iterative Tree Traversals
    • Level Order Traversal
    • Counting Nodes in a Binary Tree
    • Finding the Height of Tree
    • Finding Sum of All Nodes
  • Binary Search Tree
    • Searching in a BST
    • Inserting in a BST
    • Deleting in a BST
  • AVL Tree
    • Inserting in an AVL Tree
    • AVL Tree Rotations
    • Deleting in an AVL Tree
  • Heap
    • Inserting in a Heap
    • Deleting in a Heap
    • Heapify
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On this page
  • Count All Nodes in Binary Tree :
  • Count Nodes Having Both Children (Degree 2) :
  • Count Nodes Having One or More Children (Degree 1 or 2) :
  • Count Nodes Having Exactly One Child (Degree 1) :
  • Count Leaf Nodes (Degree 0) :
  • Alternative Way of Counting (Without Variables) :

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  1. Binary Tree

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

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Last updated 4 years ago

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