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 number of nodes in the tree. The tree is given as a collection of n pairs, where every pair represents node id and 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 tree data structure but it
can be solved as follows.

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

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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 */

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

"""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

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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 */

chevron_right



Output:

6

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

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



My Personal Notes arrow_drop_up



Article Tags :
Practice Tags :


9


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.