# https://leetcode.com/problems/flip-equivalent-binary-trees/
# 951. Flip Equivalent Binary Trees

# History:
# Google
# 1.
# Mar 12, 2020

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


# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def flipEquiv(self, root1, root2):
        """
        :type root1: TreeNode
        :type root2: TreeNode
        :rtype: bool
        """
        if not root1 or not root2:
            return not root1 and not root2

        if root1.val != root2.val:
            return False

        return (
                   self.flipEquiv(root1.left, root2.left) and
                   self.flipEquiv(root1.right, root2.right)
               ) or (
                   self.flipEquiv(root1.left, root2.right) and
                   self.flipEquiv(root1.right, root2.left)
               )
