Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree [1,2,2,3,4,4,3] is symmetric:
1 / 2 2 / \ / 3 4 4 3
But the following [1,2,2,null,3,null,3] is not:
1 / 2 2 \ 3 3
Note:
Bonus points if you could solve it both recursively and iteratively
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: bool isSymmetric(TreeNode *lhs, TreeNode *rhs) { if (NULL == lhs&&NULL == rhs)return true; if (NULL!=lhs&&NULL!=rhs&&lhs->val == rhs->val) { return isSymmetric(lhs->right, rhs->left) && isSymmetric(lhs->left, rhs->right); } return false; } bool isSymmetric(TreeNode* root) { if (!root)return true; return isSymmetric(root->left, root->right); } };
原文：http://www.cnblogs.com/csudanli/p/5877084.html