951. Flip Equivalent Binary Trees

https://leetcode.com/problems/flip-equivalent-binary-trees/

solution

class Solution:
    def flipEquiv(self, root1: Optional[TreeNode], root2: Optional[TreeNode]) -> bool:
        if root1 is None and root2 is None:
            return True

        if root1 is None or root2 is None:
            return False

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

时间复杂度：O(n) 空间复杂度：O(h)