## Algorithm

Problem Name: 951. Flip Equivalent Binary Trees

For a binary tree T, we can define a flip operation as follows: choose any node, and swap the left and right child subtrees.

A binary tree X is flip equivalent to a binary tree Y if and only if we can make X equal to Y after some number of flip operations.

Given the roots of two binary trees `root1` and `root2`, return `true` if the two trees are flip equivalent or `false` otherwise.

Example 1:

```Input: root1 = [1,2,3,4,5,6,null,null,null,7,8], root2 = [1,3,2,null,6,4,5,null,null,null,null,8,7]
Output: true
Explanation: We flipped at nodes with values 1, 3, and 5.
```

Example 2:

```Input: root1 = [], root2 = []
Output: true
```

Example 3:

```Input: root1 = [], root2 = [1]
Output: false
```

Constraints:

• The number of nodes in each tree is in the range `[0, 100]`.
• Each tree will have unique node values in the range `[0, 99]`

## Code Examples

### #1 Code Example with C++ Programming

```Code - C++ Programming```

``````
class Solution {
public:
bool flipEquiv(TreeNode* root1, TreeNode* root2) {
if (!root1 || !root2) {
return !root1 && !root2;
}
if (root1->val != root2->val) {
return false;
}
return flipEquiv(root1->left, root2->left) && flipEquiv(root1->right, root2->right)
|| flipEquiv(root1->left, root2->right) && flipEquiv(root1->right, root2->left);
}
};
``````
Copy The Code &

Input

cmd
root1 = [1,2,3,4,5,6,null,null,null,7,8], root2 = [1,3,2,null,6,4,5,null,null,null,null,8,7]

### #2 Code Example with Java Programming

```Code - Java Programming```

``````
class Solution {
public boolean flipEquiv(TreeNode root1, TreeNode root2) {
if (root1 == null && root2 == null) {
return true;
}
if (root1 == null || root2 == null) {
return false;
}
if (root1.val != root2.val) {
return false;
}
return (
(
flipEquiv(root1.left, root2.left) && flipEquiv(root1.right, root2.right)
) ||
(
flipEquiv(root1.left, root2.right) &&
flipEquiv(root1.right, root2.left)
)
);
}
}
``````
Copy The Code &

Input

cmd
root1 = [1,2,3,4,5,6,null,null,null,7,8], root2 = [1,3,2,null,6,4,5,null,null,null,null,8,7]

Output

cmd
true

### #3 Code Example with Javascript Programming

```Code - Javascript Programming```

``````
const flipEquiv = function(root1, root2) {
if(root1 == null || root2 == null) return root1 === root2
return root1.val === root2.val &&
(
(flipEquiv(root1.left, root2.left) && flipEquiv(root1.right, root2.right)) ||
(flipEquiv(root1.left, root2.right) && flipEquiv(root1.right, root2.left))
)
};
``````
Copy The Code &

Input

cmd
root1 = [], root2 = []

Output

cmd
true

### #4 Code Example with Python Programming

```Code - Python Programming```

``````
class Solution:
def flipEquiv(self, root1, root2):
if not root1 or not root2: return root1 == root2
if root1.left and root2.left and root1.left.val != root2.left.val or (not root1.left and root2.left) or (root1.left and not root2.left):
root1.left, root1.right = root1.right, root1.left
return root1.val == root2.val and self.flipEquiv(root1.left, root2.left) and self.flipEquiv(root1.right, root2.right)
``````
Copy The Code &

Input

cmd
root1 = [], root2 = []

Output

cmd
true

### #5 Code Example with C# Programming

```Code - C# Programming```

``````
namespace LeetCode
{
public class _0951_FlipEquivalentBinaryTrees
{
public bool FlipEquiv(TreeNode root1, TreeNode root2)
{
if (root1 == null) return root2 == null;
if (root2 == null) return root1 == null;
if (root1.val != root2.val) return false;

return (FlipEquiv(root1.left, root2.left) && FlipEquiv(root1.right, root2.right)) ||
(FlipEquiv(root1.left, root2.right) && FlipEquiv(root1.right, root2.left));
}
}
}
``````
Copy The Code &

Input

cmd
root1 = [], root2 = [1]

Output

cmd
false