面试题17：如何判断二叉树是否是对称二叉树？

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.

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