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 below14         // DO NOT write int main() function15         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 below14         // DO NOT write int main() function15         if (root == NULL)16             return true;17         queue
     
       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 };