101. Symmetric Tree

Well, pretty straightforward DFS solution.

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 isSymmetric(TreeNode* root) {  
13:      if (root == NULL) return true;  
14:      return helper(root->left, root->right);  
15:    }  
16:    bool helper(TreeNode *p, TreeNode *q) {  
17:      if (p == NULL && q == NULL) return true;  
18:      if (p == NULL && q != NULL || p != NULL && q == NULL || p->val != q->val) return false;  
19:      return helper(p->left, q->right) && helper(p->right, q->left);  
20:    }  
21:  };