ASMR LeetCode: Flip Equivalent Binary Trees | Chill Coding

[@LeekCodeDev]

Source Code:
# 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
class Solution:
def flipEquiv(self, root1: Optional[TreeNode], root2: Optional[TreeNode]) -> bool:
'''
To solve this problem, for each node, I'm going to compare the children for the
two roots we are given. If the left and right values are the same, or they're
just flipped, I keep recursing down the tree and return True if i reach None, because
this means that we found no inequalities. However, if the property of the children
values being the same gets broken, we return False up to the roots and we will
return False! We repeat this process to get our final True or False value, whether
we can flip the equivalent binary Tree :D
'''
if not root1 and not root2:
return True
elif not root1 or not root2:
return False
elif root1.val != root2.val:
return False
root1_left_child_val = 0 if root1.left is None else root1.left.val
root1_right_child_val = 0 if root1.right is None else root1.right.val
root2_left_child_val = 0 if root2.left is None else root2.left.val
root2_right_child_val = 0 if root2.right is None else root2.right.val
if root1_left_child_val == root2_left_child_val and root1_right_child_val == root2_right_child_val:
return self.flipEquiv(root1.left,root2.left) and self.flipEquiv(root1.right,root2.right)
elif root1_left_child_val == root2_right_child_val and root1_right_child_val == root2_left_child_val:
return self.flipEquiv(root1.left,root2.right) and self.flipEquiv(root1.right,root2.left)
else:
return False