思路:
1、两棵树p1、p2, p1与p2,p1的左与p2的右,p1的右与p2的左相同,则对称
2、递归判断全部节点
class Solution:
def isSymmetrical(self , pRoot: TreeNode) -> bool:
# write code here
if not pRoot:
return True
return self.check(pRoot, pRoot)
def check(self, p1, p2):
if not p1 and not p2:
return True
if not p1 or not p2:
return False
return p1.val == p2.val and self.check(p1.left, p2.right) and self.check(p1.right, p2.left)