Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree [1,2,2,3,4,4,3]
is symmetric:
1 / \ 2 2 / \ / \ 3 4 4 3
But the following [1,2,2,null,3,null,3]
is not:
1 / \ 2 2 \ \ 3 3
Note: Bonus points if you could solve it both recursively and iteratively.
解法
# Definition for a binary tree node. # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution: def isSymmetric(self, root: TreeNode) -> bool: if not root: return True dq = collections.deque([(root.left,root.right),]) while dq: node1, node2 = dq.popleft() #两个都是空节点直接跳过 if not node1 and not node2: continue #只要其中一个节点为空直接返回False if not node1 or not node2: return False #若两个节点都存在,那么需要判断两者的值是否相等 if node1.val != node2.val: return False # 判断左节点的左节点与右节点的右节点 # 判断左节点的右节点和右节点的左节点 dq.append((node1.left,node2.right)) dq.append((node1.right,node2.left)) return True