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# 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``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 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

 ``

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.