/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isSymmetric(TreeNode* root) {
        //树怎么能少得了递归呢？对当前节点的左右子树判断是否对称，对左右子树判断是否对称
        if(!root) return true;
        return dfs(root->left,root->right);
    }

    bool dfs(TreeNode* root1,TreeNode* root2){
        if(!root1 || !root2) return !root1 && !root2;

        return root1->val == root2->val && dfs(root1->left,root2->right) && dfs(root1->right,root2->left);
    }
};