【LEETCODE】101-Symmetric Tree

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree is symmetric:

    1

   / \

  2   2

 / \ / \

3  4 4  3

But the following is not:

    1

   / \

  2   2

   \   \

   3    3

Note:

Bonus points if you could solve it both recursively and iteratively.

confused what "{1,#,2,3}" means?

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1

  / \

 2   3

    /

   4

    \

     5

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

思路：

当 root.left == root.right 时，再去判断 root.left.left == root.right.right 和 root.left.right == root.right.left (只为记号，并不存在root.left.right之说)

然后再去递归判断 root.left.left vs root.right.right 作为root的时候，及 root.left.right vs root.right.left 作为root的时候

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    
    def compare(self,p,q):         #当两个node相等时，比较两个node作为根时的子树是否对称
        if p == q == None:
            return True
        elif p and q and p.val == q.val:        #[2,2]  
            return self.compare(p.left,q.right) and self.compare(p.right,q.left)      #[2,3,4] vs [2,4,3]
        else:
            return False
    
    def isSymmetric(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        
        if root is None:
            return True
        else:                     #[1]
            return self.compare(root.left,root.right)     #对称，需要比较两个root，所以需要写一个辅助函数