No.103 - LeetCode101 - 判断二叉树镜像 - 直接DFS即可

/**
 * 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 dfs(TreeNode* L,TreeNode* R){
        if(L->val != R->val) return false;
        bool ans = true;
        if(L->left != NULL && R->right != NULL) ans=ans&&dfs(L->left,R->right);
        if((L->left == NULL && R->right != NULL) || 
           (L->left != NULL && R->right == NULL)) return false;
        
        if(L->right != NULL && R->left != NULL) ans=ans&&dfs(L->right,R->left);
        if((L->right == NULL && R->left != NULL) || 
           (L->right != NULL && R->left == NULL)) return false;
        return ans;
    }
    bool isSymmetric(TreeNode* root) {
        if(root == NULL) return true;
        if(root->left == NULL && root->right == NULL) return true;
        if(root->left != NULL && root->right != NULL) return dfs(root->left,root->right);
        return false;
    }
};