Root of the Problem

Problem Statement :

Chef has a binary tree. The binary tree consists of 1 or more nodes. Each node has a unique integer id. Each node has up to 2 children, which are identified by their ids, and each node is the child of at most 1 other node. A node X is considered to be an ancestor of node Y if node Y is a child of node X or if there is some node Z for which X is an ancestor of Z and Y is a child of Z. No node is an ancestor of itself. A special node called the root node is an ancestor of all other nodes.

Chef has forgotten which node of his tree is the root, and wants you to help him to figure it out. Unfortunately, Chef's knowledge of the tree is incomplete. He does not remember the ids of the children of each node, but only remembers the sum of the ids of the children of each node.


Input begins with an integer T, the number of test cases. Each test case begins with an integer N, the number of nodes in the tree. N lines follow with 2 integers each: the id of a node, and the sum of the ids of its children. The second number will be 0 if the node has no children.


For each test case, output on a line a space separated list of all possible values for the id of the root node in increasing order. It is guaranteed that at least one such id exists for each test case.


1 ≤ T ≤ 50
1 ≤ N ≤ 30
All node ids are between 1 and 1000, inclusive

Sample Input

4 0
1 5
2 0
3 0
4 0
5 5
6 5

Sample Output


Solution :


                            Solution in C :

#include <stdio.h>

int main(void) {
	int t, n, id_sum, sid_sum, x, y;
	scanf("%d", &t);
	while(t--) {
	    scanf("%d", &n);
	    id_sum = 0, sid_sum = 0;
	    while(n--) {
    	    scanf("%d %d", &x, &y);
    	    id_sum += x;
    	    sid_sum += y;
	    printf("%d\n", id_sum - sid_sum);
	return 0;

                        Solution in C++ :

#include <iostream>
using namespace std;

#define ll long long int

int main() {
	ll t;
	for(ll z=1;z<=t;z++){
		ll n;
		int ans=0;
		for(int j=1;j<=n;j++){
			int x,y;
	return 0;

                        Solution in Java :

/* package codechef; // don't place package name! */

import java.util.*;
import java.lang.*;

/* Name of the class has to be "Main" only if the class is public. */
class Codechef
	public static void main (String[] args) throws java.lang.Exception
		// your code goes here
		Scanner sc = new Scanner(;
		int t = sc.nextInt();
		while(t-- > 0)
		    int n = sc.nextInt();
		    int node = 0;
		    int idsum= 0;
		    for(int i=0;i<n;i++)
		        node = node + sc.nextInt();
		        idsum = idsum + sc.nextInt();
		    int r = node - idsum;

                        Solution in Python : 
# cook your dish here
for _ in range(t):
    n = int(input())
    root = 0
    for j in range(n):
        ide,c = map(int,input().split(' '))
        root +=ide - c

View More Similar Problems

Kitty's Calculations on a Tree

Kitty has a tree, T , consisting of n nodes where each node is uniquely labeled from 1 to n . Her friend Alex gave her q sets, where each set contains k distinct nodes. Kitty needs to calculate the following expression on each set: where: { u ,v } denotes an unordered pair of nodes belonging to the set. dist(u , v) denotes the number of edges on the unique (shortest) path between nodes a

View Solution →

Is This a Binary Search Tree?

For the purposes of this challenge, we define a binary tree to be a binary search tree with the following ordering requirements: The data value of every node in a node's left subtree is less than the data value of that node. The data value of every node in a node's right subtree is greater than the data value of that node. Given the root node of a binary tree, can you determine if it's also a

View Solution →

Square-Ten Tree

The square-ten tree decomposition of an array is defined as follows: The lowest () level of the square-ten tree consists of single array elements in their natural order. The level (starting from ) of the square-ten tree consists of subsequent array subsegments of length in their natural order. Thus, the level contains subsegments of length , the level contains subsegments of length , the

View Solution →

Balanced Forest

Greg has a tree of nodes containing integer data. He wants to insert a node with some non-zero integer value somewhere into the tree. His goal is to be able to cut two edges and have the values of each of the three new trees sum to the same amount. This is called a balanced forest. Being frugal, the data value he inserts should be minimal. Determine the minimal amount that a new node can have to a

View Solution →

Jenny's Subtrees

Jenny loves experimenting with trees. Her favorite tree has n nodes connected by n - 1 edges, and each edge is ` unit in length. She wants to cut a subtree (i.e., a connected part of the original tree) of radius r from this tree by performing the following two steps: 1. Choose a node, x , from the tree. 2. Cut a subtree consisting of all nodes which are not further than r units from node x .

View Solution →

Tree Coordinates

We consider metric space to be a pair, , where is a set and such that the following conditions hold: where is the distance between points and . Let's define the product of two metric spaces, , to be such that: , where , . So, it follows logically that is also a metric space. We then define squared metric space, , to be the product of a metric space multiplied with itself: . For

View Solution →