Maximum parent children sum in Binary tree

Given a Binary Tree, find the maximum sum in a binary tree by adding the parent with its children. Exactly three Node needs to be added. If the tree does not have a node with both of its children as not NULL, return 0.

We simply traverse the tree and find the Node that has the maximum sum. We need to take care of the leaves. 

Implementation:

C++

// C++ program to find maximum sum of a node
// and its children
#include <iostream>
using namespace std;

struct Node {
    int data;
    struct Node *left, *right;
};


// insertion of Node in Tree
struct Node* newNode(int n)
{
    struct Node* root = new Node();
    root->data = n;
    root->left = root->right = NULL;
    return root;
}

int maxSum(struct Node* root)
{
    if (root == NULL)
        return 0;

    int res = maxSum(root->left);

    // if left and right link are null then
    // add all the three Node
    if (root->left != NULL && root->right != NULL) {
        int sum = root->data + root->left->data + root->right->data;
        res = max(res, sum);
    }

    return max(res, maxSum(root->right));
}

int main()
{
    struct Node* root =  newNode(15);
    root->left = newNode(16);
    root->left->left = newNode(8);
    root->left->left->left = newNode(55);
    root->left->right = newNode(67);
    root->left->right->left = newNode(44);
    root->right = newNode(17);
    root->right->left = newNode(7);
    root->right->left->right = newNode(11);
    root->right->right = newNode(41);
    cout << maxSum(root);
    return 0;
}

Java

// Java program to find
// maximum sum of a node
// and its children
import java.util.*;

// insertion of Node in Tree
class Node
{
    int data;
    Node left, right;


    Node(int key)
    {
        data = key;
        left = right = null;
    }
}
class GFG
{
    public static int maxSum(Node root)
    {
        if (root == null)
        return 0;

    int res = maxSum(root.left);

    // if left and right link are null
    // then add all the three Node
    if (root.left != null &&
        root.right != null)
    {
        int sum = root.data +
                  root.left.data +
                  root.right.data;
        res = Math.max(res, sum);
    }

    return Math.max(res, maxSum(root.right));
    }

    // Driver code
    public static void main (String[] args)
    {
        Node root = new Node(15);
        root.left = new Node(16);
        root.left.right = new Node(67);
        root.left.right.left = new Node(44);
        root.left.left = new Node(8);
        root.left.left.left = new Node(55);
        root.right = new Node(17);
        root.right.right = new Node(41);
        root.right.left = new Node(7);
        root.right.left.right = new Node(11);
        System.out.print(maxSum(root));
    }
}

// This code is contributed
// by akash1295

Python3

# Python program to find maximum
# sum of a node and its children
class newNode():

    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None

def maxSum(root):

    if (root == None):
        return 0

    res = maxSum(root.left)

    # if left and right link are None then
    # add all the three Node
    if (root.left != None and root.right != None):
        sum = root.data + root.left.data + root.right.data
        res = max(res, sum)

    return max(res, maxSum(root.right))

# Driver code
if __name__ == '__main__':
    root = newNode(15)
    root.left = newNode(16)
    root.left.left = newNode(8)
    root.left.left.left = newNode(55)
    root.left.right = newNode(67)
    root.left.right.left = newNode(44)
    root.right = newNode(17)
    root.right.left = newNode(7)
    root.right.left.right = newNode(11)
    root.right.right = newNode(41)
    print(maxSum(root))

# This code is contributed by SHUBHAMSINGH10

C#

// C# program to find
// maximum sum of a node
// and its children
using System;

// insertion of Node in Tree
public class Node
{
    public int data;
    public Node left, right;


    public Node(int key)
    {
        data = key;
        left = right = null;
    }
}
public class GFG
{
    public static int maxSum(Node root)
    {
        if (root == null)
        return 0;

    int res = maxSum(root.left);

    // if left and right link are null
    // then add all the three Node
    if (root.left != null &&
        root.right != null)
    {
        int sum = root.data +
                  root.left.data +
                  root.right.data;
        res = Math.Max(res, sum);
    }

    return Math.Max(res, maxSum(root.right));
    }

    // Driver code
    public static void Main ()
    {
        Node root = new Node(15);
        root.left = new Node(16);
        root.left.right = new Node(67);
        root.left.right.left = new Node(44);
        root.left.left = new Node(8);
        root.left.left.left = new Node(55);
        root.right = new Node(17);
        root.right.right = new Node(41);
        root.right.left = new Node(7);
        root.right.left.right = new Node(11);
        Console.Write(maxSum(root));
    }
}

/* This code is contributed PrinciRaj1992 */

Javascript

<script>

// Javascript program to find
// maximum sum of a node
// and its children

// Insertion of Node in Tree
class Node
{
    constructor(key)
    {
        this.data = key;
        this.left = null;
        this.right = null;
    }
}

function maxSum(root)
{
    if (root == null)
        return 0;

    var res = maxSum(root.left);

    // If left and right link are null
    // then add all the three Node
    if (root.left != null &&
        root.right != null)
    {
        var sum = root.data +
                  root.left.data +
                  root.right.data;
        res = Math.max(res, sum);
    }

    return Math.max(res, maxSum(root.right));
}

// Driver code
var root = new Node(15);
root.left = new Node(16);
root.left.right = new Node(67);
root.left.right.left = new Node(44);
root.left.left = new Node(8);
root.left.left.left = new Node(55);
root.right = new Node(17);
root.right.right = new Node(41);
root.right.left = new Node(7);
root.right.left.right = new Node(11);
document.write(maxSum(root));

// This code is contributed by rutvik_56

</script>

Output

91

Time Complexity: O(N) where N is the number of nodes in given binary tree.
Auxiliary Space: O(h) where h is the height of binary tree due to recursion call.