LeetCode-Symmetric Tree

树的遍历，

递归：

 1 /**
 2  * Definition for binary tree
 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 isSymmetric(TreeNode *root) {
13         // Start typing your C/C++ solution below
14         // DO NOT write int main() function
15         if (root == NULL)
16             return true;
17         return symmetric(root->left, root->right);
18     }
19     bool symmetric(TreeNode *lnode, TreeNode *rnode) {
20         if (lnode == NULL && rnode == NULL)
21             return true;
22         if (lnode == NULL || rnode == NULL) 
23             return false;
24         if (lnode->val != rnode->val) {
25             return false;
26         }
27         return symmetric(lnode->left, rnode->right) && symmetric(lnode->right, rnode->left);
28     }
29 };

迭代要用到队列；

 1 /**
 2  * Definition for binary tree
 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 isSymmetric(TreeNode *root) {
13         // Start typing your C/C++ solution below
14         // DO NOT write int main() function
15         if (root == NULL)
16             return true;
17         queue<TreeNode*> tqueue;
18         tqueue.push(root->left);
19         tqueue.push(root->right);
20         while (!tqueue.empty()) {
21             TreeNode *lnode = tqueue.front();
22             tqueue.pop();
23             TreeNode *rnode = tqueue.front();
24             tqueue.pop();
25             if (lnode == NULL && rnode == NULL)
26                 continue;
27             if (lnode == NULL || rnode == NULL || (lnode->val != rnode->val))
28                 return false;
29             tqueue.push(lnode->left);
30             tqueue.push(rnode->right);
31             tqueue.push(lnode->right);
32             tqueue.push(rnode->left);
33         }
34         return true;
35     }
36 };