Count nodes with two children at level L in a Binary Tree

Given a Binary tree, the task is to count the number of nodes with two children at a given level L.

Examples:

Input: 
          1
         /  \
        2    3
       / \    \
      4   5    6
         /    / \
        7    8   9
L = 2
Output: 1

Input:
          20
         /   \
        8     22
       / \    / \
      5   3  4   25
     / \  / \     \
    1  10 2  14    6
L = 3
Output: 2

Approach: Initialize a variable count = 0. Recursively traverse the tree in a level order manner. If the current level is same as the given level, then check whether the current node has two children. If it has two children then increment the variable count.



Below is the implementation of the above approach:

C++

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// C++ program to find number of full nodes
// at a given level
#include <bits/stdc++.h>
using namespace std;
  
// A binary tree node
struct Node {
    int data;
    struct Node *left, *right;
};
  
// Utility function to allocate memory for a new node
struct Node* newNode(int data)
{
    struct Node* node = new (struct Node);
    node->data = data;
    node->left = node->right = NULL;
    return (node);
}
  
// Function that returns the height of binary tree
int height(struct Node* root)
{
    if (root == NULL)
        return 0;
  
    int lheight = height(root->left);
    int rheight = height(root->right);
  
    return max(lheight, rheight) + 1;
}
  
// Level Order traversal to find the number of nodes
// having two children
void LevelOrder(struct Node* root, int level, int& count)
{
    if (root == NULL)
        return;
  
    if (level == 1 && root->left && root->right)
        count++;
  
    else if (level > 1) {
        LevelOrder(root->left, level - 1, count);
        LevelOrder(root->right, level - 1, count);
    }
}
  
// Returns the number of full nodes
// at a given level
int CountFullNodes(struct Node* root, int L)
{
    // Stores height of tree
    int h = height(root);
  
    // Stores count of nodes at a given level
    // that have two children
    int count = 0;
  
    LevelOrder(root, L, count);
  
    return count;
}
  
// Driver code
int main()
{
    struct Node* root = newNode(7);
    root->left = newNode(5);
    root->right = newNode(6);
    root->left->left = newNode(8);
    root->left->right = newNode(1);
    root->left->left->left = newNode(2);
    root->left->left->right = newNode(11);
    root->right->left = newNode(3);
    root->right->right = newNode(9);
    root->right->right->right = newNode(13);
    root->right->right->left = newNode(10);
    root->right->right->right->left = newNode(4);
    root->right->right->right->right = newNode(12);
  
    int L = 3;
  
    cout << CountFullNodes(root, L);
  
    return 0;
}

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Java

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// Java program to find number of full nodes 
// at a given level 
class GFG
{
  
//INT class
static class INT
{
    int a;
}
  
// A binary tree node 
static class Node 
    int data; 
    Node left, right; 
}; 
  
// Utility function to allocate memory for a new node 
static Node newNode(int data) 
    Node node = new Node(); 
    node.data = data; 
    node.left = node.right = null
    return (node); 
  
// Function that returns the height of binary tree 
static int height(Node root) 
    if (root == null
        return 0
  
    int lheight = height(root.left); 
    int rheight = height(root.right); 
  
    return Math.max(lheight, rheight) + 1
  
// Level Order traversal to find the number of nodes 
// having two children 
static void LevelOrder( Node root, int level, INT count) 
    if (root == null
        return
  
    if (level == 1 && root.left!=null && root.right!=null
        count.a++; 
  
    else if (level > 1)
    
        LevelOrder(root.left, level - 1, count); 
        LevelOrder(root.right, level - 1, count); 
    
  
// Returns the number of full nodes 
// at a given level 
static int CountFullNodes( Node root, int L) 
    // Stores height of tree 
    int h = height(root); 
  
    // Stores count of nodes at a given level 
    // that have two children 
    INT count =new INT();
    count.a = 0
  
    LevelOrder(root, L, count); 
  
    return count.a; 
  
// Driver code 
public static void main(String args[])
    Node root = newNode(7); 
    root.left = newNode(5); 
    root.right = newNode(6); 
    root.left.left = newNode(8); 
    root.left.right = newNode(1); 
    root.left.left.left = newNode(2); 
    root.left.left.right = newNode(11); 
    root.right.left = newNode(3); 
    root.right.right = newNode(9); 
    root.right.right.right = newNode(13); 
    root.right.right.left = newNode(10); 
    root.right.right.right.left = newNode(4); 
    root.right.right.right.right = newNode(12); 
  
    int L = 3
  
    System.out.print( CountFullNodes(root, L)); 
  
}
  
// This code is contributed by Arnab Kundu

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C#

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// C# program to find number of full nodes 
// at a given level
using System;
  
class GFG 
  
// INT class 
public class INT 
    public int a; 
  
// A binary tree node 
public class Node 
    public int data; 
    public Node left, right; 
}; 
  
// Utility function to allocate memory for a new node 
static Node newNode(int data) 
    Node node = new Node(); 
    node.data = data; 
    node.left = node.right = null
    return (node); 
  
// Function that returns the height of binary tree 
static int height(Node root) 
    if (root == null
        return 0; 
  
    int lheight = height(root.left); 
    int rheight = height(root.right); 
  
    return Math.Max(lheight, rheight) + 1; 
  
// Level Order traversal to find the number of nodes 
// having two children 
static void LevelOrder( Node root, int level, INT count) 
    if (root == null
        return
  
    if (level == 1 && root.left!=null && root.right!=null
        count.a++; 
  
    else if (level > 1) 
    
        LevelOrder(root.left, level - 1, count); 
        LevelOrder(root.right, level - 1, count); 
    
  
// Returns the number of full nodes 
// at a given level 
static int CountFullNodes( Node root, int L) 
    // Stores height of tree 
    int h = height(root); 
  
    // Stores count of nodes at a given level 
    // that have two children 
    INT count =new INT(); 
    count.a = 0; 
  
    LevelOrder(root, L, count); 
  
    return count.a; 
  
// Driver code 
public static void Main(String []args) 
    Node root = newNode(7); 
    root.left = newNode(5); 
    root.right = newNode(6); 
    root.left.left = newNode(8); 
    root.left.right = newNode(1); 
    root.left.left.left = newNode(2); 
    root.left.left.right = newNode(11); 
    root.right.left = newNode(3); 
    root.right.right = newNode(9); 
    root.right.right.right = newNode(13); 
    root.right.right.left = newNode(10); 
    root.right.right.right.left = newNode(4); 
    root.right.right.right.right = newNode(12); 
  
    int L = 3; 
  
    Console.Write( CountFullNodes(root, L)); 
  
  
// This code has been contributed by 29AjayKumar

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Output:

2


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