LeetCode101. Symmetric Tree

题目描述：

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

解题思路： 

判断是否是一个对称树，简单，采用Divide And Conquer的方法去做，把大的问题不断分成小问题取解决。

Python代码： 

class Solution:
    def isSymmetric(self, root: 'TreeNode') -> 'bool':
        if not root:
            return True
        return self.isMirror(root.left, root.right)
    def isMirror(self, left, right): 
        if left == None and right == None:
            return True
        elif left != None and right != None:
            if left.val == right.val:
                return self.isMirror(left.left, right.right) and self.isMirror(left.right, right.left)
            else:
                return False
        else:
            return False

这里在做题的时候，遇到了个小问题，python中的&是按位与，优先级比==， !=要高，而and是逻辑与，比==要低，因此在使用&时，&两边的不等式判断要加括号，而and则不需要。