"""
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.

Write a function that determines whether two binary trees are flip equivalent.  The trees are given by root nodes root1 and root2.


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.
         1
       /  \
      2    3
     / \   /
    4  5  6
      / \
     7   8

          1
       /   \
      3     2
       \   / \
       6  4   5
             / \
            8   7
Note:

Each tree will have at most 100 nodes.
Each value in each tree will be a unique integer in the range [0, 99].
"""
# notes:

# TODO iteratively + complexity
class SolutionT951(object):
    def flipEquiv(self, root1, root2):
        """
        :type root1: TreeNode
        :type root2: TreeNode
        :rtype: bool
        """
        if not root1 and not root2: return True
        if not root1 or not root2: return False
        left1 = self.flipEquiv(root1.left, root2.left)
        left2 = self.flipEquiv(root1.left, root2.right)
        right1 = self.flipEquiv(root1.right, root2.right)
        right2 = self.flipEquiv(root1.right, root2.left)
        return root1.val == root2.val and ((left1 and right1) or (left2 and right2))

c00efc117803e5032fe8a22e0a2bb266483454e6
<mask token>

<mask token>


class SolutionT951(object):
    <mask token>

<mask token>


class SolutionT951(object):

    def flipEquiv(self, root1, root2):
        """
        :type root1: TreeNode
        :type root2: TreeNode
        :rtype: bool
        """
        if not root1 and not root2:
            return True
        if not root1 or not root2:
            return False
        left1 = self.flipEquiv(root1.left, root2.left)
        left2 = self.flipEquiv(root1.left, root2.right)
        right1 = self.flipEquiv(root1.right, root2.right)
        right2 = self.flipEquiv(root1.right, root2.left)
        return root1.val == root2.val and (left1 and right1 or left2 and right2
            )

"""
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.

Write a function that determines whether two binary trees are flip equivalent.  The trees are given by root nodes root1 and root2.


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.
         1
       /  \
      2    3
     / \   /
    4  5  6
      / \
     7   8

          1
       /   \
      3     2
       \   / \
       6  4   5
             / \
            8   7
Note:

Each tree will have at most 100 nodes.
Each value in each tree will be a unique integer in the range [0, 99].
"""
# notes:

# TODO iteratively + complexity
class SolutionT951(object):
    def flipEquiv(self, root1, root2):
        """
        :type root1: TreeNode
        :type root2: TreeNode
        :rtype: bool
        """
        if not root1 and not root2: return True
        if not root1 or not root2: return False
        left1 = self.flipEquiv(root1.left, root2.left)
        left2 = self.flipEquiv(root1.left, root2.right)
        right1 = self.flipEquiv(root1.right, root2.right)
        right2 = self.flipEquiv(root1.right, root2.left)
        return root1.val == root2.val and ((left1 and right1) or (left2 and right2))
