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 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? > read more on how binary tree is serialized on OJ.
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}".
1 /**
2 * Definition for binary tree
3 * struct TreeNode {
4 * int val;
5 * TreeNode *left;
6 * TreeNode *right;
7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {}
8 * };
9 */
代码：
1 class Solution {
2 public:
3 bool isSymmetric(TreeNode *root) {
4 return isSymmetricCore(root,root);
5
6 }
7 bool isSymmetricCore(TreeNode* root1,TreeNode* root2){
8 if(root1 == nullptr && root2 == nullptr)
9 return true;
10 if(root1 == nullptr || root2 == nullptr)
11 return false;
12 if(root1->val != root2->val)
13 return false;
14 return isSymmetricCore(root1->left,root2->right) && isSymmetricCore(root1->right,root2->left);
15
16 }
17 };
原文：https://www.cnblogs.com/zjuhaohaoxuexi/p/11778994.html