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+# 951. Flip Equivalent Binary Trees
+
+"""
+for a binarytree 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 equivalent.
+"""
+
+
+# Definition for a binary tree node.
+class TreeNode(object):
+ def __init__(self, val=0, left=None, right=None):
+ self.val = val
+ self.left = left
+ self.right = right
+
+
+class Solution(object):
+ def flipEquiv(self, root1, root2):
+ """
+ :type root1: Optional[TreeNode]
+ :type root2: Optional[TreeNode]
+ :rtype: bool
+ """
+ s1, s2 = [root1], [root2]
+ while s1 and s2:
+ n1, n2 = s1.pop(), s2.pop()
+ if n1 is None or n2 is None:
+ continue
+ elif not n1 or not n1 or n1.val != n2.val:
+ return False
+ if (
+ n1.left is None
+ or n2.left is None
+ or n1.left
+ and n2.left
+ and n1.left.val == n2.left.val
+ ):
+ s1.extend([n1.left, n1.right])
+ else:
+ s1.extend([n1.right, n1.left])
+ s2.extend([n2.left, n2.right])
+ return not s1 and not s2
+
+
+if __name__ == "__main__":
+ obj = Solution()
+ print(
+ obj.flipEquiv(
+ root1=[1, 2, 3, 4, 5, 6, None, None, None, 7, 8],
+ root2=[1, 3, 2, None, 6, 4, 5, None, None, None, None, 8, 7],
+ )
+ )
+ print(obj.flipEquiv(root1=[], root2=[]))
+ print(obj.flipEquiv(root1=[], root2=[1]))