题解 | #对称的二叉树#

struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
    TreeNode(int x) :
            val(x), left(NULL), right(NULL) {
    }
};
*/
class Solution {
public:
    bool isSymmetrical(TreeNode* root) {
        //即是判断左子树的右节点是否与右子树的左节点是否相等
        if(!root) return true;
        return dfs(root->left, root->right);
    }
    
    bool dfs(TreeNode* p1, TreeNode* p2){
        //当p1或者p2为空时，需要两者均为空时才返回true，否则返回false
        if(!p1 || !p2) return !p1 && !p2;
        if(p1->val != p2->val) return false;
        return dfs(p1->left, p2->right) && dfs(p1->right, p2->left);
    }
    

};