题目：

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}".

解题思路：采用先序遍历的思路对根节点的两颗子树进行遍历，在遍历的时候需要注意，左子树的遍历过程中，先访左子树的左儿子，然后访问左子树的右儿子；右子树的遍历过程中，先访问右子树的右儿子，再访问右子树的左儿子。即镜像的对左右两颗子树进行比较。


递归代码：

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isSymmetric(TreeNode *root) {
        TreeNode *LeftSubTree=root,*RightSubTree=root;
        return PreorderTraverse(LeftSubTree,RightSubTree);
    }
    
private:
    bool PreorderTraverse(TreeNode *LeftSubTree,TreeNode *RightSubTree){
        if((LeftSubTree==nullptr)&&(RightSubTree==nullptr))return true;
        if((LeftSubTree==nullptr)||(RightSubTree==nullptr))return false;
        return (LeftSubTree->val==RightSubTree->val)&&PreorderTraverse(LeftSubTree->left,RightSubTree->right)&&PreorderTraverse(LeftSubTree->right,RightSubTree->left);
    }
};

迭代代码：

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isSymmetric(TreeNode *root) {
        stack<TreeNode *> s;
        if(root==nullptr)return true;
        s.push(root->left);
        s.push(root->right);
        while(!s.empty()){
            TreeNode *Right=s.top();s.pop();
            TreeNode *Left=s.top();s.pop();
            
            if(Left==nullptr&&Right==nullptr)continue;
            if(Left==nullptr||Right==nullptr)return false;
            
            if(Left->val==Right->val){
                s.push(Left->left);
                s.push(Right->right);
                s.push(Left->right);
                s.push(Right->left);
            }else{
                return false;
            }       
        }
        return true;
    }
};