题解 | #对称的二叉树#
对称的二叉树
https://www.nowcoder.com/practice/ff05d44dfdb04e1d83bdbdab320efbcb
# class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None # # 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可 # # # @param pRoot TreeNode类 # @return bool布尔型 # class Solution: def isSymmetrical(self , pRoot: TreeNode) -> bool: # write code here if not pRoot: return True q = [] if pRoot.left: q.append(pRoot.left) if pRoot.right: q.append(pRoot.right) while q: if len(q) % 2 != 0: return False n = len(q) left = [] right = [] left_node = [] right_node = [] for _ in range(n//2): node_l = q.pop(0) if node_l: left.append(node_l.val) left_node.append(node_l) else: left.append(node_l) for _ in range(n // 2): node_r = q.pop() if node_r: right_node.append(node_r) right.append(node_r.val) else: right.append(node_r) if left == right: for node in left_node: q.append(node.left) q.append(node.right) for node in right_node[::-1]: q.append(node.left) q.append(node.right) else: return False return True