101. 对称二叉树(深搜)

迭代是不可能迭代的

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     bool flag = true; //判断标志
13     bool isSymmetric(TreeNode* root) {
14         if (!root) return 1;
15         dfs(root->left, root->right);
16         return flag;
17     }
18     void dfs(TreeNode* L, TreeNode* R) {
19         if (!flag || !L && !R) return; //剪枝
20         else if (!L || !R || L->val != R->val) { //判断不对称
21             flag = false;
22             return;
23         }
24         dfs(L->left, R->right); //对称搜索比较
25         dfs(L->right, R->left);    
26     }
27 };