leetcode-solution/Python/0951-flip-equivalent-binary-trees.py at main · hogan-tech/leetcode-solution · GitHub

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# time complexity: O(n)
# space complexity: O(n)
from typing import Optional


class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right


def traverse(node):
    if node is None:
        return
    if node.val:
        print(node.val)
    if node.left:
        traverse(node.left)
    if node.right:
        traverse(node.right)


class Solution:
    def flipEquiv(self, root1: Optional[TreeNode], root2: Optional[TreeNode]) -> bool:
        if not root1 and not root2:
            return True
        if not root1 or not root2:
            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.left, root2.right) and self.flipEquiv(root1.right, root2.left))


root1 = TreeNode(1)
root1.left = TreeNode(2)
root1.right = TreeNode(3)
root1.left.left = TreeNode(4)
root1.left.right = TreeNode(5)
root1.right.left = TreeNode(6)
root1.left.right.left = TreeNode(7)
root1.left.right.right = TreeNode(8)


root2 = TreeNode(1)
root2.left = TreeNode(3)
root2.right = TreeNode(2)
root2.left.right = TreeNode(6)
root2.right.left = TreeNode(4)
root2.right.right = TreeNode(5)
root2.right.right.left = TreeNode(8)
root2.right.right.right = TreeNode(7)
print(Solution().flipEquiv(root1, root2))