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题目：
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree is symmetric:
1 / 2 2 / \ / 3 4 4 3
But the following is not:
1 / 2 2 \ 3 3
Note:
Bonus points if you could solve it both recursively and iteratively.
分类：Tree BFS DFS
代码：
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 13 bool isSymmetric(TreeNode *root) { 14 if (!root) return true; 15 return helper(root->left, root->right); 16 } 17 18 bool helper(TreeNode* p, TreeNode* q) { 19 if (!p && !q) { 20 return true; 21 } else if (!p || !q) { 22 return false; 23 } 24 25 if (p->val != q->val) { 26 return false; 27 } 28 29 return helper(p->left,q->right) && helper(p->right, q->left); 30 } 31 };
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原文地址：http://www.cnblogs.com/zhangbaochong/p/5492988.html
