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? > read more on how binary tree is serialized on OJ.

分析：这道题使用递归，判断左右子结点是否相等，然后咱判断pLeft->left和pRight->right、pLeft->right和pRight->left结点是否相同。

/**
 * 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 check(TreeNode *pLeft,TreeNode *pRight)
    {
        if(pLeft==NULL&&pRight==NULL)
            return true;
        if(pLeft==NULL || pRight==NULL)
            return false;
        return pLeft->val==pRight->val && check(pLeft->left,pRight->right) && check(pLeft->right,pRight->left);
    }
    bool isSymmetric(TreeNode *root) {
        if(root==NULL)
            return true;
        return check(root->left,root->right);
    }
};

这道题一开始使用中序遍历，然后不能Accept，出现在{1,2,3,3,#,2,#}不能过，中序遍历是有问题的，不应该这样判断。中序遍历的二叉树不是唯一的，所以不能确定二叉树的顺序。