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) {
    }
};
*/
/*以下情况表示为镜像：
1.左右子树都为空
2.左子树的左根==右子树的右根
3.左子树的右根==右子树的左根
*/
class Solution {
public:
    bool myisSymmetrical(TreeNode* ltree, TreeNode* rtree){
        if(!ltree && !rtree){
            return true;
        }
        if(!ltree || !rtree){
            return false;
        }
        return ltree->val == rtree->val && myisSymmetrical(ltree->right, rtree->left) && myisSymmetrical(ltree->left, rtree->right);
    }
    bool isSymmetrical(TreeNode* pRoot) {
        if(!pRoot){
            return true;
        }
        return myisSymmetrical(pRoot->left, pRoot->right);
    }

};