Node having maximum sum of immediate children and itself in n-ary tree

Given an N-Ary tree, find and return the node for which sum of data of all children and the node itself is maximum. In the sum, data of node itself and data of its immediate children is to be taken. 
For example in the given tree,

maxSum Node = 4 with maximum sum of 28

The idea is we will maintain a integer variable maxsum which contains the maximum sum yet, and a resnode node pointer which points to the node with maximum sum. 
Traverse the tree and maintain the sum of root and data of all its immediate children in currsum
integer variable and update the maxsum variable accordingly. 

Implementation:

C++

// CPP program to find the node whose children
// and node sum is maximum.
#include <bits/stdc++.h>
using namespace std;

// Structure of a node of an n-ary tree
struct Node {
    int key;
    vector<Node*> child;
};

// Utility function to create a new tree node
Node* newNode(int key)
{
    Node* temp = new Node;
    temp->key = key;
    return temp;
}

// Helper function to find the node
void maxSumUtil(Node* root, Node** resNode,
                int* maxsum)
{
    // Base Case
    if (root == NULL)
        return;

    // curr contains the sum of the root and
    // its children
    int currsum = root->key;

    // total no of children
    int count = root->child.size();

    // for every child call recursively
    for (int i = 0; i < count; i++) {
        currsum += root->child[i]->key;
        maxSumUtil(root->child[i], resNode, maxsum);
    }

    // if curr is greater than sum, update it
    if (currsum > *maxsum) {

        // resultant node
        *resNode = root;
        *maxsum = currsum;
    }
    return;
}

// Function to find the node having max sum of
// children and node
int maxSum(Node* root)
{
    // resultant node with max sum of children
    // and node
    Node* resNode;

    // sum of node and its children
    int maxsum = 0;

    maxSumUtil(root, &resNode, &maxsum);

    // return the key of resultant node
    return resNode->key;
}

// Driver program
int main()
{
    /*   Let us create below tree
    *              1
    *          /   |  \
    *         2   3   4
    *        / \    / |  \ \
    *       5   6  7  8  9  10
    */

    Node* root = newNode(1);
    (root->child).push_back(newNode(2));
    (root->child).push_back(newNode(3));
    (root->child).push_back(newNode(4));
    (root->child[0]->child).push_back(newNode(5));
    (root->child[0]->child).push_back(newNode(6));
    (root->child[2]->child).push_back(newNode(5));
    (root->child[2]->child).push_back(newNode(6));
    (root->child[2]->child).push_back(newNode(6));

    cout << maxSum(root) << endl;

    return 0;
}

Java

// Java program to find the node whose children
// and node sum is maximum.
import java.util.*;

class GFG
{

// Structure of a node of an n-ary tree
static class Node
{
    int key;
    Vector<Node> child;
    Node()
    {
        child = new Vector<Node>();
    }
};

// Utility function to create a new tree node
static Node newNode(int key)
{
    Node temp = new Node();
    temp.key = key;
    return temp;
}

static int maxsum;

// resultant node with max sum of children
// and node
static Node resNode;

// Helper function to find the node
static void maxSumUtil(Node root)
{
    // Base Case
    if (root == null)
        return;

    // curr contains the sum of the root and
    // its children
    int currsum = root.key;

    // total no of children
    int count = root.child.size();

    // for every child call recursively
    for (int i = 0; i < count; i++)
    {
        currsum += root.child.get(i).key;
        maxSumUtil(root.child.get(i));
    }

    // if curr is greater than sum, update it
    if (currsum > maxsum)
    {

        // resultant node
        resNode = root;
        maxsum = currsum;
    }
    return;
}

// Function to find the node having max sum of
// children and node
static int maxSum(Node root)
{

    // sum of node and its children
    int maxsum = 0;

    maxSumUtil(root);

    // return the key of resultant node
    return resNode.key;
}

// Driver code
public static void main(String args[])
{
    /* Let us create below tree
                1
            / | \
            2 3 4
        / \ / | \ \
        5 6 7 8 9 10
    */

    Node root = newNode(1);
    (root.child).add(newNode(2));
    (root.child).add(newNode(3));
    (root.child).add(newNode(4));
    (root.child.get(0).child).add(newNode(5));
    (root.child.get(0).child).add(newNode(6));
    (root.child.get(2).child).add(newNode(5));
    (root.child.get(2).child).add(newNode(6));
    (root.child.get(2).child).add(newNode(6));

    System.out.print( maxSum(root) );
}
}

// This code is contributed by Arnab Kundu

Python3

# Python3 program to find the node
# whose children and node sum is maximum.

# Structure of a node of an n-ary tree
class Node:

    def __init__(self, key):
        self.key = key
        self.child = []

# Helper function to find the node
def maxSumUtil(root, resNode, maxsum):

    # Base Case
    if root == None:
        return

    # curr contains the sum of the root
    # and its children
    currsum = root.key

    # total no of children
    count = len(root.child)

    # for every child call recursively
    for i in range(0, count):
        currsum += root.child[i].key
        resNode, maxsum = maxSumUtil(root.child[i],
                                     resNode, maxsum)

    # if curr is greater than sum,
    # update it
    if currsum > maxsum:

        # resultant node
        resNode = root
        maxsum = currsum

    return resNode, maxsum

# Function to find the node having
# max sum of children and node
def maxSum(root):

    # resultant node with max
    # sum of children and node
    resNode, maxsum = Node(None), 0
    resNode, maxsum = maxSumUtil(root, resNode,
                                       maxsum)

    # return the key of resultant node
    return resNode.key

# Driver Code
if __name__ == "__main__":

    root = Node(1)
    (root.child).append(Node(2))
    (root.child).append(Node(3))
    (root.child).append(Node(4))
    (root.child[0].child).append(Node(5))
    (root.child[0].child).append(Node(6))
    (root.child[2].child).append(Node(5))
    (root.child[2].child).append(Node(6))
    (root.child[2].child).append(Node(6))

    print(maxSum(root))

# This code is contributed by Rituraj Jain

C#

// C# program to find the node whose children
// and node sum is maximum
using System;
using System.Collections.Generic;

class GFG
{

// Structure of a node of an n-ary tree
public class Node
{
    public int key;
    public List<Node> child;
    public Node()
    {
        child = new List<Node>();
    }
};

// Utility function to create a new tree node
static Node newNode(int key)
{
    Node temp = new Node();
    temp.key = key;
    return temp;
}

static int maxsum;

// resultant node with max sum of children
// and node
static Node resNode;

// Helper function to find the node
static void maxSumUtil(Node root)
{
    // Base Case
    if (root == null)
        return;

    // curr contains the sum of the root and
    // its children
    int currsum = root.key;

    // total no of children
    int count = root.child.Count;

    // for every child call recursively
    for (int i = 0; i < count; i++)
    {
        currsum += root.child[i].key;
        maxSumUtil(root.child[i]);
    }

    // if curr is greater than sum, update it
    if (currsum > maxsum)
    {

        // resultant node
        resNode = root;
        maxsum = currsum;
    }
    return;
}

// Function to find the node having max sum of
// children and node
static int maxSum(Node root)
{

    // sum of node and its children
    int maxsum = 0;

    maxSumUtil(root);

    // return the key of resultant node
    return resNode.key;
}

// Driver code
public static void Main(String []args)
{
    /* Let us create below tree
                1
            / | \
            2 3 4
        / \ / | \ \
        5 6 7 8 9 10
    */

    Node root = newNode(1);
    (root.child).Add(newNode(2));
    (root.child).Add(newNode(3));
    (root.child).Add(newNode(4));
    (root.child[0].child).Add(newNode(5));
    (root.child[0].child).Add(newNode(6));
    (root.child[2].child).Add(newNode(5));
    (root.child[2].child).Add(newNode(6));
    (root.child[2].child).Add(newNode(6));

    Console.Write( maxSum(root) );
}
}

// This code has been contributed by 29AjayKumar

Javascript

<script>

// JavaScript program to find the node whose children
// and node sum is maximum


// Structure of a node of an n-ary tree
class Node
{
    constructor()
    {
        this.key = null;
        this.child = []
    }
};

// Utility function to create a new tree node
function newNode(key)
{
    var temp = new Node();
    temp.key = key;
    return temp;
}

var maxsum = 0;

// resultant node with max sum of children
// and node
var resNode = 0;

// Helper function to find the node
function maxSumUtil(root)
{
    // Base Case
    if (root == null)
        return;

    // curr contains the sum of the root and
    // its children
    var currsum = root.key;

    // total no of children
    var count = root.child.length;

    // for every child call recursively
    for (var i = 0; i < count; i++)
    {
        currsum += root.child[i].key;
        maxSumUtil(root.child[i]);
    }

    // if curr is greater than sum, update it
    if (currsum > maxsum)
    {

        // resultant node
        resNode = root;
        maxsum = currsum;
    }
    return;
}

// Function to find the node having max sum of
// children and node
function maxSum(root)
{

    // sum of node and its children
    var maxsum = 0;

    maxSumUtil(root);

    // return the key of resultant node
    return resNode.key;
}

// Driver code
/* Let us create below tree
            1
        / | \
        2 3 4
    / \ / | \ \
    5 6 7 8 9 10
*/
var root = newNode(1);
(root.child).push(newNode(2));
(root.child).push(newNode(3));
(root.child).push(newNode(4));
(root.child[0].child).push(newNode(5));
(root.child[0].child).push(newNode(6));
(root.child[2].child).push(newNode(5));
(root.child[2].child).push(newNode(6));
(root.child[2].child).push(newNode(6));
document.write( maxSum(root) );

</script>

Output

4

Time complexity: O(n) where n is the number of nodes in the binary tree.
Auxiliary Space: O(h) where h is the height of the binary tree.