uva 839 Not so Mobile

原题：
Before being an ubiquous communications gadget, a mobile was just a structure made of strings and wires suspending colourfull things. This kind of mobile is usually found hanging
over cradles of small babies.The figure illustrates a simple mobile. It is just a wire,suspended by a string, with an object on each side. It can also be seen as a kind of lever with the fulcrum on the point where the string ties the wire. From the lever principle we know that to balance a simple mobile the product of the weight of the objects by their distance to the fulcrum must be equal. That is W l ×D l = W r ×D r where D l is the left distance,D r is the right distance, W l is the left weight and W r is the right weight.In a more complex mobile the object may be replaced by a sub-mobile, as shown in the next figure.In this case it is not so straightforward to check if the mobile is balanced so we need you to write a program that, given a description of a mobile as input, checks whether the mobile is in equilibrium or not.
Input
The input begins with a single positive integer on a line by itself indicating the number
of the cases following, each of them as described below. This line is followed by a blank
line, and there is also a blank line between two consecutive inputs.
The input is composed of several lines, each containing 4 integers separated by a single space.
The 4 integers represent the distances of each object to the fulcrum and their weights, in the format:
W l D l W r D r
If W l or W r is zero then there is a sub-mobile hanging from that end and the following lines define the the sub-mobile. In this case we compute the weight of the sub-mobile as the sum of weights of all its objects, disregarding the weight of the wires and strings. If both W l and W r are zero then the following lines define two sub-mobiles: first the left then the right one.
Output
For each test case, the output must follow the description below. The outputs of two
consecutive cases will be separated by a blank line.
Write ‘YES’ if the mobile is in equilibrium, write ‘NO’ otherwise.
Sample Input
1
0 2 0 4
0 3 0 1
1 1 1 1
2 4 4 2
1 6 3 2
Sample Output
YES
中文：
给你一个树形的天平，让你判断是否这个树上所有的天平是否平衡。如果其中一个力臂为0，那表示这个力臂上的砝码是一个天平。
最后结果不要换行。

#include <bits/stdc++.h>using namespace std;int ans;int dfs(){    int wl,dl,wr,dr;    cin>>wl>>dl>>wr>>dr;    if(!wl)        wl=dfs();    if(!wr)        wr=dfs();    if(wl*dl!=wr*dr)        ans=false;    return wl+wr;}int main(){    ios::sync_with_stdio(false);    int n;    cin>>n;    while(n--)    {        ans=true;        dfs();        if(ans)            cout<<"YES"<<endl;        else            cout<<"NO"<<endl;        if(n)            cout<<endl;    }    return 0;}

解答：
二叉树求和遍历，然后判断是否平衡即可。