Count half nodes in a Binary tree (Iterative and Recursive) - GeeksforGeeks

28 Jan, 2022
Given A binary Tree, how do you count all the half nodes (which has only one child) without using recursion? Note leaves should not be touched as they have both children as NULL. 
Input : Root of below tree
Output : 3
Nodes 7, 5 and 9 are half nodes as one of 
their child is Null. So count of half nodes
in the above tree is 3
Iterative The idea is to use level-order traversal to solve this problem efficiently. 
1) Create an empty Queue Node and push root node to Queue.
2) Do following while nodeQeue is not empty.
   a) Pop an item from Queue and process it.
      a.1) If it is half node then increment count++.
   b) Push left child of popped item to Queue, if available.
   c) Push right child of popped item to Queue, if available.
Below is the implementation of this idea.
C++
Java
Python3
C#
Javascript
// C++ program to count half nodes in a Binary Tree#include <bits/stdc++.h>using namespace std; // A binary tree Node has data, pointer to left// child and a pointer to right childstruct Node{    int data;    struct Node* left, *right;}; // Function to get the count of half Nodes in// a binary treeunsigned int gethalfCount(struct Node* node){    // If tree is empty    if (!node)        return 0;     int count = 0; // Initialize count of half nodes     // Do level order traversal starting from root    queue<Node *> q;    q.push(node);    while (!q.empty())    {        struct Node *temp = q.front();        q.pop();         if (!temp->left && temp->right ||            temp->left && !temp->right)            count++;         if (temp->left != NULL)            q.push(temp->left);        if (temp->right != NULL)            q.push(temp->right);    }    return count;} /* Helper function that allocates a new   Node with the given data and NULL left   and right pointers. */struct Node* newNode(int data){    struct Node* node = new Node;    node->data = data;    node->left = node->right = NULL;    return (node);} // Driver programint main(void){    /* 2     / \    7     5    \     \     6     9    / \ /    1 11 4    Let us create Binary Tree shown in    above example */     struct Node *root = newNode(2);    root->left     = newNode(7);    root->right     = newNode(5);    root->left->right = newNode(6);    root->left->right->left = newNode(1);    root->left->right->right = newNode(11);    root->right->right = newNode(9);    root->right->right->left = newNode(4);     cout << gethalfCount(root);     return 0;}
// Java program to count half nodes in a Binary Tree// using Iterative approachimport java.util.Queue;import java.util.LinkedList; // Class to represent Tree nodeclass Node{    int data;    Node left, right;     public Node(int item)    {        data = item;        left = null;        right = null;    }} // Class to count half nodes of Treeclass BinaryTree{     Node root;     /* Function to get the count of half Nodes in    a binary tree*/    int gethalfCount()    {        // If tree is empty        if (root==null)            return 0;         // Do level order traversal starting from root        Queue<Node> queue = new LinkedList<Node>();        queue.add(root);         int count=0; // Initialize count of half nodes        while (!queue.isEmpty())        {             Node temp = queue.poll();            if (temp.left!=null && temp.right==null ||                temp.left==null && temp.right!=null)                count++;             // Enqueue left child            if (temp.left != null)                queue.add(temp.left);             // Enqueue right child            if (temp.right != null)                queue.add(temp.right);        }        return count;    }     public static void main(String args[])    {        /* 2          / \        7     5        \     \        6     9        / \ /        1 11 4        Let us create Binary Tree shown in        above example */        BinaryTree tree_level = new BinaryTree();        tree_level.root = new Node(2);        tree_level.root.left = new Node(7);        tree_level.root.right = new Node(5);        tree_level.root.left.right = new Node(6);        tree_level.root.left.right.left = new Node(1);        tree_level.root.left.right.right = new Node(11);        tree_level.root.right.right = new Node(9);        tree_level.root.right.right.left = new Node(4);         System.out.println(tree_level.gethalfCount());     }}
# Python program to count# half nodes in a Binary Tree# using iterative approach # A node structureclass Node:     # A utility function to create a new node    def __init__(self ,key):        self.data = key        self.left = None        self.right = None # Iterative Method to count half nodes of binary treedef gethalfCount(root):     # Base Case    if root is None:        return 0     # Create an empty queue for level order traversal    queue = []     # Enqueue Root and initialize count    queue.append(root)     count = 0 #initialize count for half nodes    while(len(queue) > 0):         node = queue.pop(0)         # if it is half node then increment count        if node.left is not None and node.right is None or node.left is None and node.right is not None:            count = count+1         #Enqueue left child        if node.left is not None:            queue.append(node.left)         # Enqueue right child        if node.right is not None:            queue.append(node.right)     return count #Driver Program to test above function root = Node(2)root.left = Node(7)root.right = Node(5)root.left.right = Node(6)root.left.right.left = Node(1)root.left.right.right = Node(11)root.right.right = Node(9)root.right.right.left = Node(4)  print "%d" %(gethalfCount(root))
// C# program to count half nodes in a Binary Tree// using Iterative approachusing System;using System.Collections.Generic; // Class to represent Tree nodepublic class Node{    public int data;    public Node left, right;     public Node(int item)    {        data = item;        left = null;        right = null;    }} // Class to count half nodes of Treepublic class BinaryTree{     Node root;     /* Function to get the count of half Nodes in    a binary tree*/    int gethalfCount()    {        // If tree is empty        if (root == null)            return 0;         // Do level order traversal starting from root        Queue<Node> queue = new Queue<Node>();        queue.Enqueue(root);         int count = 0; // Initialize count of half nodes        while (queue.Count != 0)        {             Node temp = queue.Dequeue();            if (temp.left != null && temp.right == null ||                temp.left == null && temp.right != null)                count++;             // Enqueue left child            if (temp.left != null)                queue.Enqueue(temp.left);             // Enqueue right child            if (temp.right != null)                queue.Enqueue(temp.right);        }        return count;    }     // Driver code    public static void Main()    {        /* 2        / \        7 5        \ \        6 9        / \ /        1 11 4        Let us create Binary Tree shown in        above example */        BinaryTree tree_level = new BinaryTree();        tree_level.root = new Node(2);        tree_level.root.left = new Node(7);        tree_level.root.right = new Node(5);        tree_level.root.left.right = new Node(6);        tree_level.root.left.right.left = new Node(1);        tree_level.root.left.right.right = new Node(11);        tree_level.root.right.right = new Node(9);        tree_level.root.right.right.left = new Node(4);         Console.WriteLine(tree_level.gethalfCount());    }} // This code contributed by Rajput-Ji
<script> // Javascript program to count half nodes in a Binary Tree// using Iterative approach // Class to represent Tree nodeclass Node{    constructor(item)    {        this.data = item;        this.left = null;        this.right = null;    }} // Class to count half nodes of Treevar root; /* Function to get the count of half Nodes ina binary tree*/function gethalfCount(){     // If tree is empty    if (root == null)        return 0;             // Do level order traversal starting from root    var queue = [];    queue.push(root);    var count = 0; // Initialize count of half nodes    while (queue.length != 0)    {        var temp = queue.shift();        if (temp.left != null && temp.right == null ||            temp.left == null && temp.right != null)            count++;                     // push left child        if (temp.left != null)            queue.push(temp.left);                     // push right child        if (temp.right != null)            queue.push(temp.right);    }    return count;} // Driver code/* 2/ \7 5\ \6 9/ \ /1 11 4Let us create Binary Tree shown inabove example */root = new Node(2);root.left = new Node(7);root.right = new Node(5);root.left.right = new Node(6);root.left.right.left = new Node(1);root.left.right.right = new Node(11);root.right.right = new Node(9);root.right.right.left = new Node(4);document.write(gethalfCount()); // This code is contributed by famously.</script>
Output: 
 3
Time Complexity: O(n) Auxiliary Space: O(n) where, n is number of nodes in given binary tree
Recursive The idea is to traverse the tree in postorder. If current node is half, we increment result by 1 and add returned values of left and right subtrees. 
C++
Java
Python3
C#
Javascript
// C++ program to count half nodes in a Binary Tree#include <bits/stdc++.h>using namespace std; // A binary tree Node has data, pointer to left// child and a pointer to right childstruct Node{    int data;    struct Node* left, *right;}; // Function to get the count of half Nodes in// a binary treeunsigned int gethalfCount(struct Node* root){    if (root == NULL)       return 0;     int res = 0;    if  ((root->left == NULL && root->right != NULL) ||         (root->left != NULL && root->right == NULL))       res++;     res += (gethalfCount(root->left) + gethalfCount(root->right));    return res;} /* Helper function that allocates a new   Node with the given data and NULL left   and right pointers. */struct Node* newNode(int data){    struct Node* node = new Node;    node->data = data;    node->left = node->right = NULL;    return (node);} // Driver programint main(void){    /* 2     / \    7     5    \     \     6     9    / \ /    1 11 4    Let us create Binary Tree shown in    above example */     struct Node *root = newNode(2);    root->left     = newNode(7);    root->right     = newNode(5);    root->left->right = newNode(6);    root->left->right->left = newNode(1);    root->left->right->right = newNode(11);    root->right->right = newNode(9);    root->right->right->left = newNode(4);     cout << gethalfCount(root);     return 0;}
// Java program to count half nodes in a Binary Treeimport java.util.*;class GfG { // A binary tree Node has data, pointer to left// child and a pointer to right childstatic class Node{    int data;    Node left, right;} // Function to get the count of half Nodes in// a binary treestatic int gethalfCount(Node root){    if (root == null)    return 0;     int res = 0;    if ((root.left == null && root.right != null) ||        (root.left != null && root.right == null))    res++;     res += (gethalfCount(root.left)            + gethalfCount(root.right));    return res;} /* Helper function that allocates a newNode with the given data and NULL leftand right pointers. */static Node newNode(int data){    Node node = new Node();    node.data = data;    node.left = null;    node.right = null;    return (node);} // Driver programpublic static void main(String[] args){    /* 2    / \    7 5    \ \    6 9    / \ /    1 11 4    Let us create Binary Tree shown in    above example */     Node root = newNode(2);    root.left = newNode(7);    root.right = newNode(5);    root.left.right = newNode(6);    root.left.right.left = newNode(1);    root.left.right.right = newNode(11);    root.right.right = newNode(9);    root.right.right.left = newNode(4);     System.out.println(gethalfCount(root)); }}
# Python program to count half nodes in a binary tree # A node structureclass newNode:    def __init__(self, data):        self.data = data        self.left = self.right = None # Function to get the count of half Nodes in a binary treedef gethalfCount(root):    if root == None:        return 0    res = 0    if(root.left == None and root.right != None) or \        (root.left != None and root.right == None):        res += 1    res += (gethalfCount(root.left) + \            gethalfCount(root.right))    return res # Driver program''' 2    / \    7     5    \     \    6     9    / \ /    1 11 4    Let us create Binary Tree shown in    above example ''' root = newNode(2)root.left = newNode(7)root.right = newNode(5)root.left.right = newNode(6)root.left.right.left = newNode(1)root.left.right.right = newNode(11)root.right.right = newNode(9)root.right.right.left = newNode(4)print(gethalfCount(root)) # This code is contributed by simranjenny84
// C# program to count half nodes in a Binary Treeusing System; class GfG{     // A binary tree Node has data, pointer to left    // child and a pointer to right child    public class Node    {        public int data;        public Node left, right;    }     // Function to get the count of half Nodes in    // a binary tree    static int gethalfCount(Node root)    {        if (root == null)        return 0;         int res = 0;        if ((root.left == null && root.right != null) ||            (root.left != null && root.right == null))        res++;         res += (gethalfCount(root.left)                + gethalfCount(root.right));        return res;    }     /* Helper function that allocates a new    Node with the given data and NULL left    and right pointers. */    static Node newNode(int data)    {        Node node = new Node();        node.data = data;        node.left = null;        node.right = null;        return (node);    }     // Driver code    public static void Main()    {        /* 2        / \        7 5        \ \        6 9        / \ /        1 11 4        Let us create Binary Tree shown in        above example */         Node root = newNode(2);        root.left = newNode(7);        root.right = newNode(5);        root.left.right = newNode(6);        root.left.right.left = newNode(1);        root.left.right.right = newNode(11);        root.right.right = newNode(9);        root.right.right.left = newNode(4);         Console.WriteLine(gethalfCount(root));    }} /* This code contributed by PrinciRaj1992 */
<script> // Javascript program to count half// nodes in a Binary Treeclass Node{    constructor(data)    {        this.left = null;        this.right = null;        this.data = data;    }} // Function to get the count of half// Nodes in a binary treefunction gethalfCount(root){    if (root == null)        return 0;     let res = 0;         if ((root.left == null && root.right != null) ||        (root.left != null && root.right == null))        res++;     res += (gethalfCount(root.left) +            gethalfCount(root.right));    return res;} /* Helper function that allocates a newNode with the given data and NULL leftand right pointers. */function newNode(data){    let node = new Node(data);    return (node);} // Driver code/*          2        / \        7  5        \   \         6    9        / \  /       1  11 4Let us create Binary Tree shown inabove example */let root = newNode(2);root.left = newNode(7);root.right = newNode(5);root.left.right = newNode(6);root.left.right.left = newNode(1);root.left.right.right = newNode(11);root.right.right = newNode(9);root.right.right.left = newNode(4); document.write(gethalfCount(root)); // This code is contributed by divyeshrabadiya07 </script>
Output : 
3
Time Complexity: O(n) Auxiliary Space: O(n) where, n is number of nodes in given binary tree
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