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.
confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.
DFS。
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 check(TreeNode* left,TreeNode* right) 13 { 14 if(!left && !right) return true; 15 if(!left || !right) return false; 16 return (left->val==right->val)&&check(left->left,right->right)&&check(left->right,right->left); 17 } 18 bool isSymmetric(TreeNode* root) { 19 if(!root) return true; 20 return check(root->left,root->right); 21 } 22 };
原文：http://www.cnblogs.com/Sean-le/p/4805952.html