Question

您好，我有一个代码来检查树是否为mirror tree。但是，它并没有通过leetcode上的所有测试用例。

我在线上与其他来源进行了核对，但似乎无法弄清楚代码中的错误。

# 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):
        """
        :type root: TreeNode
        :rtype: bool
        """
        if root is None:
            return True

        if root.left and root.right:
            return self.is_symmetric(root.left,root.right) 

    def is_symmetric(self, left, right):

        if (left == None and right == None):
            return True
        if right is None or left is None:
            return False
        if left.val == right.val:
            return True
        return self.is_symmetric(left.left,right.right) and self.is_symmetric(left.right,right.left)

谢谢！

Answer 1

当检查root.left and root.right的条件为False时，您的isSymmetric函数隐式返回None。相反，它应该继续检查root.left或root.right中的一个不是None，在这种情况下，它应该返回False；否则root.left和root.right均为None，并且应返回True。另外，在is_symmetric函数中，仅因为left.val == right.val并不意味着左子树与右子树对称，因为它们之下可能有更多子代。当False时，它应该简单地返回left.val != right.val：

class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

class Solution:
    def isSymmetric(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        if root is None:
            return True
        if root.left and root.right:
            return self.is_symmetric(root.left,root.right)
        if root.left or root.right:
            return False
        return True

    def is_symmetric(self, left, right):

        if left is None and right is None:
            return True
        if right is None or left is None or left.val != right.val:
            return False
        return self.is_symmetric(left.left, right.right) and self.is_symmetric(left.right, right.left)

Answer 2

检查树是否为镜像树的另一种方法是找到每个节点的路径，然后检查所有路径是否反向：

#include <iostream>

int main() {
    std::cout << removeAccents("ᾤκεον") << std::endl;
}

输出：

class Tree:
  def __init__(self, **kwargs):
    self.__dict__ = {i:kwargs.get(i) for i in ['value', 'left', 'right']}
  @staticmethod
  def is_reversed(a, b):
     return True if not a and not b else all(int(c) == (not int(d)) for c, d in zip(a, b))
  def get_paths(self, path = ''):
    yield [self.value, path]
    yield from getattr(self.right, 'get_paths', lambda _:[])(path+'1')
    yield from getattr(self.left, 'get_paths', lambda _:[])(path+'0')

t = Tree(value=31, left=Tree(value=16, left=Tree(value=7), right=Tree(value=24, right=Tree(value=29), left=Tree(value=19))), right=Tree(value=45))
t1 = Tree(value=31, right=Tree(value=16, right=Tree(value=7), left=Tree(value=24, right=Tree(value=19), left=Tree(value=29))), left=Tree(value=45))
"""
        31     |      31
      /    \   |     /  \
    16      45 |   45   16
   /  \        |        / \
  7   24       |       24  7
     /  \      |      /  \
    19   29    |     29  19

"""
_paths1 = dict(list(t.get_paths()))
_paths2 = dict(list(t1.get_paths()))
print(all(Tree.is_reversed(_paths1[i], _paths2[i]) for i in _paths1))

检查树是否是一面镜子

2 个答案: