题目浅析

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  • 简单地说,就是判断两个二叉树,对其中一个二叉树做任意次对任意节点的翻转操作,能否使两个二叉树相同。

思路分享

代码解答(强烈建议自行解答后再看)

  • 参考题解
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# 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:
if not root1 or not root2:
return root1 is root2
return root1.val == root2.val and ((self.flipEquiv(root1.left, root2.right) and self.flipEquiv(root1.right, root2.left)) or (self.flipEquiv(root1.left, root2.left) and self.flipEquiv(root1.right, root2.right)))