Python recursive DFS solution with comments
\n# Definition for a binary tree node.\n# class TreeNode:\n#     def __init__(self, val=0, left=None, right=None):\n#         self.val = val\n#         self.lef
```\n# Definition for a binary tree node.\n# class TreeNode:\n#     def __init__(self, val=0, left=None, right=None):\n#         self.val = val\n#         self.left = left\n#         self.right = right\nclass Solution:\n    def flipEquiv(self, root1: TreeNode, root2: TreeNode) -> bool:\n        \n        \'\'\'\n        We think of root1 as the "match" tree, that we are looking to match \n        by flipping parts of root2, the "flipper".\n        \n        We want to traverse the match, while trying a flip and a non-flip at each node in the flipper.\n        This makes sure we try every combination of flips, and exit early when they don\'t match.\n        Recursively do this.\n        If the regular version matches or the flipped version matches, we return true.\n        \'\'\'\n        \n        def dfs(match, flipper):\n            \n            # Both are None.\n            if not match and not flipper:\n                return True\n            \n            # Only one is None, so they don\'t match.\n            if not (match and flipper):\n                return False\n            \n            # They both exist, but values don\'t match.\n            if match.val != flipper.val:\n                return False\n            \n            # No flip.\n            regular = dfs(match.left, flipper.left) and dfs(match.right, flipper.right)\n            \n            # Flip.\n            flipped = dfs(match.left, flipper.right) and dfs(match.right, flipper.left)\n            \n            return regular or flipped\n        \n        return dfs(root1, root2)\n```
['Depth-First Search', 'Recursion', 'Python', 'Python3']