Find root of the tree where children id sum for every node is given

Consider a binary tree whose nodes have ids from 1 to n where n is the number of nodes in the tree. The tree is given as a collection of n pairs, where every pair represents node id and the sum of children ids.

Examples:

Input : 1 5
        2 0
        3 0
        4 0
        5 5
        6 5
Output: 6
Explanation: In this case, two trees can 
be made as follows and 6 is the root node.
   6          6
   \         / \
    5       1   4
   / \       \
  1   4       5
 / \         / \
2   3       2   3

Input : 4 0
Output: 4
Explanation: Clearly 4 does 
not have any children and is the
only node i.e., the root node.

At first sight, this question appears to be a typical question of a tree data structure but it 
can be solved as follows.

Every node id appears in children sum except root. So if we do the sum of all ids and subtract it from the sum of all children’s sums, we get the root.

Implementation:

C++

// Find root of tree where children
// sum for every node id is given.
#include<bits/stdc++.h>
using namespace std;

int findRoot(pair<int, int> arr[], int n)
{
   // Every node appears once as an id, and
   // every node except for the root appears
   // once in a sum.  So if we subtract all
   // the sums from all the ids, we're left
   // with the root id.
   int root = 0;
   for (int i=0; i<n; i++)
     root += (arr[i].first - arr[i].second);

   return root;
}

// Driver code
int main()
{
    pair<int, int> arr[] = {{1, 5}, {2, 0},
           {3, 0}, {4, 0}, {5, 5}, {6, 5}};
    int n = sizeof(arr)/sizeof(arr[0]);
    printf("%d\n", findRoot(arr, n));
    return 0;
}

Java

// Find root of tree where children
// sum for every node id is given.

class GFG
{

    static class pair
    {

        int first, second;

        public pair(int first, int second)
        {
            this.first = first;
            this.second = second;
        }

    }

    static int findRoot(pair arr[], int n)
    {
        // Every node appears once as an id, and
        // every node except for the root appears
        // once in a sum. So if we subtract all
        // the sums from all the ids, we're left
        // with the root id.
        int root = 0;
        for (int i = 0; i < n; i++)
        {
            root += (arr[i].first - arr[i].second);
        }

        return root;
    }

    // Driver code
    public static void main(String[] args)
    {
        pair arr[] = {new pair(1, 5), new pair(2, 0),
                    new pair(3, 0), new pair(4, 0),
                    new pair(5, 5), new pair(6, 5)};
        int n = arr.length;
        System.out.printf("%d\n", findRoot(arr, n));
    }

}

/* This code is contributed by PrinciRaj1992 */

Python3

"""Find root of tree where children
sum for every node id is given"""

def findRoot(arr, n) :

    # Every node appears once as an id, and
    # every node except for the root appears
    # once in a sum. So if we subtract all
    # the sums from all the ids, we're left
    # with the root id.
    root = 0
    for i in range(n):
        root += (arr[i][0] - arr[i][1])
    return root

# Driver Code
if __name__ == '__main__':

    arr = [[1, 5], [2, 0],
           [3, 0], [4, 0],
           [5, 5], [6, 5]]
    n = len(arr)
    print(findRoot(arr, n))

# This code is contributed
# by SHUBHAMSINGH10

C#

// C# Find root of tree where children
// sum for every node id is given.
using System;

class GFG
{

    public class pair
    {

        public int first, second;

        public pair(int first, int second)
        {
            this.first = first;
            this.second = second;
        }

    }

    static int findRoot(pair []arr, int n)
    {
        // Every node appears once as an id, and
        // every node except for the root appears
        // once in a sum. So if we subtract all
        // the sums from all the ids, we're left
        // with the root id.
        int root = 0;
        for (int i = 0; i < n; i++)
        {
            root += (arr[i].first - arr[i].second);
        }

        return root;
    }

    // Driver code
    public static void Main(String[] args)
    {
        pair []arr = {new pair(1, 5), new pair(2, 0),
                    new pair(3, 0), new pair(4, 0),
                    new pair(5, 5), new pair(6, 5)};
        int n = arr.Length;
        Console.Write("{0}\n", findRoot(arr, n));
    }

}

/* This code is contributed by PrinciRaj1992 */

Javascript

<script>

// Javascript: Find root of tree where children
// sum for every node id is given.

/*Every node appears once as an id, and
every node except for the root appears
once in a sum. So if we subtract all
the sums from all the ids, we're left
with the root id. */


const findRoot = (nodeList) => {
    let root = 0;
    nodeList.forEach(element =>
    root+=(Number(element[0]) - Number(element[1])));
    return root ;
}


let nodeList = [[1, 5], [2, 0],
                [3, 0], [4, 0],
                [5, 5], [6, 5]];
let root = findRoot(nodeList);
document.write(root);

// This code is contributed by bharathmkulkarni

</script>

Output

6

Time Complexity: O(n), Where n is the length of the given array.
Auxiliary Space: O(1), As no extra space is used.

This article is contributed by Sunidhi Chaudhary. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.