flipEquiv.py

"""
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.
Given the roots of two binary trees root1 and root2, return true if the two trees are flip equivalent or false otherwise.

Example 1:
Flipped Trees Diagram
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.

Example 2:
Input: root1 = [], root2 = []

Output: true
Example 3:
Input: root1 = [], root2 = [1]
Output: false
 
Constraints:
The number of nodes in each tree is in the range [0, 100].
Each tree will have unique node values in the range [0, 99].
"""
from typing import Optional
# Definition for a binary tree node.
class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

    def __str__(self): 
        return f'[{self.val}, {self.left}, {self.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))
        


# Test Cases
s = Solution()
print(s.flipEquiv(TreeNode(1, TreeNode(2, TreeNode(4), TreeNode(5, TreeNode(7), TreeNode(8))), TreeNode(3, TreeNode(6))), TreeNode(1, TreeNode(3, TreeNode(6)), TreeNode(2, TreeNode(4), TreeNode(5, TreeNode(8), TreeNode(7))))) == True)
print(s.flipEquiv(None, None) == True)
print(s.flipEquiv(None, TreeNode(1)) == False)