Count nodes with sum of path made by only left child nodes at least K

Last Updated : 30 Sep, 2022

Given a binary tree and an integer K, the task is to write a program to count the number of nodes such that the path from the current node to a leaf consisting of only the left child of nodes has a sum greater than or equal to K.

Examples:

Input: K = 15,
Tree:
8
/ \
9 10
/ \ / \
11 12 13 7
/ \ / / / \
6 9 6 7 15 11

Output: 4
Explanation: Nodes with left children sum ? 'K' are :
Node 8 = 9 + 11 + 6 = 26
Node 9 = 11 + 6 = 17
Node 10 = 13 + 7 = 20
Node 7 = 15

Input: K = 20,
Tree:
3
/ \
16 10

Output: 0
Explanation: No nodes with left children sum ? 20

Approach: To solve the problem follow the below idea:

One simple approach is to find the sum of left children nodes for every node and compare it with K.

Follow the steps to solve the problem:

Maintain counter with name count
Implement a recursive function leftNodeSum that recursively calculates the sum of the left nodes for every node in the tree
Use a recursive function to traverse through every node
Check the sum with K, if it is greater than K then increment the count.
Return the count as the answer.

Below is the implementation of the above approach.

C++

// C++ code for the above approach:

#include <bits/stdc++.h>
using namespace std;

// Node structure
struct node {
    int data;
    struct node* left;
    struct node* right;
    node(int x)
    {
        this->data = x;
        this->left = this->right = NULL;
    }
};

// Function to count left node sum
int leftNodeSum(struct node* head)
{
    if (!head)
        return 0;
    return head->data + leftNodeSum(head->left);
}

// Function to traverse the tree
int traverseNodes(struct node* head, int k, int* count)
{
    if (!head)
        return 0;

    int sum = leftNodeSum(head->left);
    if (sum >= k)
        (*count)++;

    traverseNodes(head->left, k, count);
    traverseNodes(head->right, k, count);
}

// Driver code
int main()
{
    int K = 15;
    int count = 0;
    struct node* start = NULL;
    start = new node(8);
    start->left = new node(9);
    start->right = new node(10);
    start->left->left = new node(11);
    start->left->right = new node(12);
    start->right->left = new node(13);
    start->right->right = new node(7);
    start->left->left->left = new node(6);
    start->left->left->right = new node(9);
    start->left->right->left = new node(6);
    start->right->left->left = new node(7);
    start->right->right->left = new node(15);
    start->right->right->right = new node(11);

    // Function call
    traverseNodes(start, K, &count);

    cout << count << endl;
    return 0;
}

// C code to implement the approach

#include <stdio.h>
#include <stdlib.h>

// Structure of a tree node
struct node {
    int data;
    struct node* left;
    struct node* right;
};
int count = 0;

// Function to add a new node
struct node* addNode(int data)
{
    struct node* n
        = (struct node*)malloc(sizeof(struct node));
    n->data = data;
    n->left = NULL;
    n->right = NULL;
    return n;
}

// Function to find the sum of left children
int leftNodeSum(struct node* head)
{
    if (!head)
        return 0;
    return head->data + leftNodeSum(head->left);
}

// Function to traverse the tree
int traverseNodes(struct node* head, int k)
{
    if (!head)
        return 0;

    int sum = leftNodeSum(head->left);
    if (sum >= k)
        count++;

    traverseNodes(head->left, k);
    traverseNodes(head->right, k);
}

// Driver code
void main()
{
    int K = 15;
    struct node* start = NULL;
    start = addNode(8);
    start->left = addNode(9);
    start->right = addNode(10);
    start->left->left = addNode(11);
    start->left->right = addNode(12);
    start->right->left = addNode(13);
    start->right->right = addNode(7);
    start->left->left->left = addNode(6);
    start->left->left->right = addNode(9);
    start->left->right->left = addNode(6);
    start->right->left->left = addNode(7);
    start->right->right->left = addNode(15);
    start->right->right->right = addNode(11);

    // Function call
    traverseNodes(start, K);

    printf("%d", count);
}

Java

// Java code for the above approach:
import java.io.*;

// Node structure
class node {
    int data;
    node left, right;
    node(int x)
    {
        this.data = x;
        this.left = this.right = null;
    }
}

class GFG {

    // Function to count left node sum
    public static int leftNodeSum(node head)
    {
        if (head == null)
            return 0;
        return head.data + leftNodeSum(head.left);
    }

    // Function to traverse the tree
    public static void traverseNodes(node head, int k,
                                     int count[])
    {
        if (head == null)
            return;

        int sum = leftNodeSum(head.left);
        if (sum >= k)
            count[0]++;

        traverseNodes(head.left, k, count);
        traverseNodes(head.right, k, count);
    }

    // Driver Code
    public static void main(String[] args)
    {
        int K = 15;
        int count[] = new int[1];
        count[0] = 0;
        node start = new node(8);
        start.left = new node(9);
        start.right = new node(10);
        start.left.left = new node(11);
        start.left.right = new node(12);
        start.right.left = new node(13);
        start.right.right = new node(7);
        start.left.left.left = new node(6);
        start.left.left.right = new node(9);
        start.left.right.left = new node(6);
        start.right.left.left = new node(7);
        start.right.right.left = new node(15);
        start.right.right.right = new node(11);

        // Function call
        traverseNodes(start, K, count);

        System.out.println(count[0]);
    }
}

// This code is contributed by Rohit Pradhan

Python3

# Python3 code to implement the approach

# Class to define the structure of a tree node
class node:
    def __init__(self):
        self.data = None;
        self.left = None;
        self.right = None;

count = 0

# Function to add a new node
def addNode(data):
    n = node()
    n.data = data
    n.left = None
    n.right = None
    return n

# Function to find the sum of left children
def leftNodeSum(head):
    if not head:
        return 0
    return head.data + leftNodeSum(head.left)

# Function to traverse the tree
def traverseNodes(head, k):
    global count
    if not head:
        return 0
    sum = leftNodeSum(head.left)
    if sum >= k:
        count+= 1
    traverseNodes(head.left, k)
    traverseNodes(head.right, k)

# Driver code
if __name__ == '__main__':
    K = 15
    start = None
    start = addNode(8)
    start.left = addNode(9)
    start.right = addNode(10)
    start.left.left = addNode(11)
    start.left.right = addNode(12)
    start.right.left = addNode(13)
    start.right.right = addNode(7)
    start.left.left.left = addNode(6)
    start.left.left.right = addNode(9)
    start.left.right.left = addNode(6)
    start.right.left.left = addNode(7)
    start.right.right.left = addNode(15)
    start.right.right.right = addNode(11)
    
    # Function call
    traverseNodes(start, K)
    print(count)

// Include namespace system
using System;

// Node structure
public class node
{
    public int data;
    public node left;
    public node right;
    public node(int x)
    {
        this.data = x;
        this.left = this.right = null;
    }
}
public class GFG
{
  
    // Function to count left node sum
    public static int leftNodeSum(node head)
    {
        if (head == null)
        {
            return 0;
        }
        return head.data + GFG.leftNodeSum(head.left);
    }
  
    // Function to traverse the tree
    public static void traverseNodes(node head, int k, int[] count)
    {
        if (head == null)
        {
            return;
        }
        var sum = GFG.leftNodeSum(head.left);
        if (sum >= k)
        {
            count[0]++;
        }
        GFG.traverseNodes(head.left, k, count);
        GFG.traverseNodes(head.right, k, count);
    }
  
    // Driver Code
    public static void Main(String[] args)
    {
        var K = 15;
        int[] count = new int[1];
        count[0] = 0;
        var start = new node(8);
        start.left = new node(9);
        start.right = new node(10);
        start.left.left = new node(11);
        start.left.right = new node(12);
        start.right.left = new node(13);
        start.right.right = new node(7);
        start.left.left.left = new node(6);
        start.left.left.right = new node(9);
        start.left.right.left = new node(6);
        start.right.left.left = new node(7);
        start.right.right.left = new node(15);
        start.right.right.right = new node(11);
      
        // Function call
        GFG.traverseNodes(start, K, count);
        Console.WriteLine(count[0]);
    }
}

// This code is contributed by aadityaburujwale.

JavaScript

class node {
constructor(){
let data;
let left;
let right;
}

};
let count = 0;

function addNode(data){
let n = new node();
n.data = data;
n.left = null;
n.right = null;
return n;
}

function leftNodeSum(head){
if(!head)return 0;
return head.data + leftNodeSum(head.left);
}

function traverseNodes(head, k){
if(!head)return 0;

let sum = leftNodeSum(head.left);
if(sum >= k)count++;

traverseNodes(head.left, k);
traverseNodes(head.right, k);

}


let k = 15;
let start = null;
start = addNode(8);
start.left = addNode(9);
start.right = addNode(10);
start.left.left = addNode(11);
start.left.right = addNode(12);
start.right.left = addNode(13);
start.right.right = addNode(7);
start.left.left.left = addNode(6);
start.left.left.right = addNode(9);
start.left.right.left = addNode(6);
start.right.left.left = addNode(7);
start.right.right.left = addNode(15);
start.right.right.right = addNode(11);


traverseNodes(start, k);

console.log(count);

Output

Time Complexity: O(N^2)
Auxiliary Space: O(N) the space used by the recursion stack

Check if a Binary Tree consists of a pair of leaf nodes with sum K

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Count nodes with sum of path made by only left child nodes at least K

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