题目:原题链接(简单)
标签:树、二叉树、深度优先搜索
解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
Ans 1 (Python) | O(N) | O(N) | 44ms (74.25%) |
Ans 2 (Python) | |||
Ans 3 (Python) |
解法一:
class Solution:
def isSymmetric(self, root: TreeNode) -> bool:
def dfs(node1, node2):
if node1 and node2:
return node1.val == node2.val and dfs(node1.right, node2.left) and dfs(node1.left, node2.right)
elif node1 or node2:
return False
else:
return True
return dfs(root.left, root.right) if root else True