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

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++

// 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 child
struct Node
{
    int data;
    struct Node* left, *right;
};
  
// Function to get the count of half Nodes in
// a binary tree
unsigned 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 program
int 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

// Java program to count half nodes in a Binary Tree
// using Iterative approach
import java.util.Queue;
import java.util.LinkedList;
  
// Class to represent Tree node
class Node
{
    int data;
    Node left, right;
  
    public Node(int item)
    {
        data = item;
        left = null;
        right = null;
    }
}
  
// Class to count half nodes of Tree
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 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

   
# Python program to count
# half nodes in a Binary Tree
# using iterative approach
  
# A node structure
class 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 tree
def 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#

// C# program to count half nodes in a Binary Tree 
// using Iterative approach 
using System;
using System.Collections.Generic; 
  
// Class to represent Tree node 
public 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 Tree 
public 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


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++

// 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 child
struct Node
{
    int data;
    struct Node* left, *right;
};
  
// Function to get the count of half Nodes in
// a binary tree
unsigned 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 program
int 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

// Java program to count half nodes in a Binary Tree 
import java.util.*;
class GfG {
  
// A binary tree Node has data, pointer to left 
// child and a pointer to right child 
static class Node 
    int data; 
    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 program 
public 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)); 
  
}

C#

// C# program to count half nodes in a Binary Tree 
using 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 */


Output :

3

Time Complexity: O(n)
Auxiliary Space: O(n)
where, n is number of nodes in given binary tree

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This article is attributed to GeeksforGeeks.org

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