## 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 equivelent 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
```

Example 4:

```Input: root1 = [0,null,1], root2 = []
Output: false
```

Example 5:

```Input: root1 = [0,null,1], root2 = [0,1]
Output: true
```

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]`.

## Rust Solution

``````struct Solution;
use rustgym_util::*;

trait FlipEq {
fn flip_eq(root1: &TreeLink, root2: &TreeLink) -> bool;
}

impl FlipEq for TreeLink {
fn flip_eq(root1: &TreeLink, root2: &TreeLink) -> bool {
if let (Some(node1), Some(node2)) = (root1, root2) {
let val1 = node1.borrow().val;
let val2 = node2.borrow().val;
let left1 = &node1.borrow().left;
let right1 = &node1.borrow().right;
let left2 = &node2.borrow().left;
let right2 = &node2.borrow().right;
val1 == val2
&& (Self::flip_eq(left1, left2) && Self::flip_eq(right1, right2)
|| Self::flip_eq(left1, right2) && Self::flip_eq(right1, left2))
} else {
root1 == root2
}
}
}

impl Solution {
fn flip_equiv(root1: TreeLink, root2: TreeLink) -> bool {
}
}

#[test]
fn test() {
let root1 = tree!(
1,
tree!(2, tree!(4), tree!(5, tree!(7), tree!(8))),
tree!(3, tree!(6), None)
);
let root2 = tree!(
1,
tree!(3, None, tree!(6)),
tree!(2, tree!(4), tree!(5, tree!(8), tree!(7)))
);
let res = true;
assert_eq!(Solution::flip_equiv(root1, root2), res);
let root1 = tree!(0, tree!(3), tree!(1, None, tree!(2)));
let root2 = tree!(0, tree!(3, tree!(2), None), tree!(1));
let res = false;
assert_eq!(Solution::flip_equiv(root1, root2), res);
}
``````

Having problems with this solution? Click here to submit an issue on github.