题解 | #牛群的轴对称结构#

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param root TreeNode类 
     * @return bool布尔型
     */
    bool isSymmetric_2(TreeNode* root_1, TreeNode* root_2)
    {
        if(!root_1 && !root_2)
            return true;
        
        if((!root_1 || !root_2) || (root_1->val != root_2->val))
            return false;

        return isSymmetric_2(root_1->left, root_2->right) && isSymmetric_2(root_1->right, root_2->left);
    }

    bool isSymmetric(TreeNode* root) {
        // write code here
        if(!root)
            return true;

        return isSymmetric_2(root->left, root->right);
    }
};