Example
1 1
/ \ / \
2 3 3 2
Output: 1
1 1
/ \ / \
2 3 2 3
Output: 0
Note: you may assume that root of both the given tree as 1.
For each test case in a new line print 1 if both the trees are mirrors of each other else print 0.
For Input:
2
3 2
1 2 1 3
1 3 1 2
7 7
1 2 1 3 2 4 2 5 5 6 6 7 6 9
1 3 1 2 2 5 2 4 5 6 6 9 6 7
Your Output is:
1
1
#include<bits/stdc++.h>
using namespace std;
int main() {
//code
int t;
cin >> t;
while (t--) {
int n, e, x, y;
cin >> n >> e;
stack < int > s[e + 1];
for (int i = 0; i < e; ++i) {
cin >> x >> y;
s[x].push(y);
//pushing the child correspond to parent.
}
int res = 1;
for (int i = 0; i < e; ++i) {
cin >> x >> y;
if (res == 1)
//when top of stack correspond to parent same to y. bcz stack work on LIFO.
// then pop the top element correspond to that parent and res =1;
if (s[x].top() == y) {
s[x].pop();
res = 1;
} else {
// if not equal to y that mean there is no mirror
res = 0;
}
}
if (res)
cout << 1 << endl;
else
cout << 0 << endl;
}
return 0;
}
