**Twin Trees - Amazon Top Interview Questions**

### Problem Statement :

Given two binary trees, root0 and root1, return whether their structure and values are equal. Constraints n ≤ 100,000 where n is the number of nodes in root0 m ≤ 100,000 where m is the number of nodes in root1 Example 1 Input root0 = [0, [5, null, null], [9, null, null]] root1 = [0, [5, null, null], [9, null, null]] Output True Explanation These two trees have the same values and same structure. Example 2 Input root0 = [0, [5, null, null], [9, null, null]] root1 = [1, [2, null, null], [3, null, null]] Output False Explanation These two trees are not twins since their values are different. Example 3 Input root0 = [0, [5, null, null], null] root1 = [0, null, [5, null, null]] Output False Explanation These two trees are not twins since their structure is different.

### Solution :

` ````
Solution in C++ :
bool solve(Tree* root0, Tree* root1) {
if (root0 == NULL and root1 == NULL)
return true;
else if (root0 == NULL ^ root1 == NULL)
return false;
return (root0->val == root1->val and solve(root0->left, root1->left) and
solve(root0->right, root1->right));
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public boolean solve(Tree root0, Tree root1) {
if (root0 == null && root1 == null) {
return true;
} else if (root0 == null ^ root1 == null) {
return false;
}
return root0.val == root1.val && solve(root0.left, root1.left)
&& solve(root0.right, root1.right);
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, root0, root1):
if not root0 or not root1:
return not root0 and not root1
if root0.val != root1.val:
return False
return self.solve(root0.left, root1.left) and self.solve(root0.right, root1.right)
```

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