标签:
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.
题目描述:
对称树的判断。
代码实现:
/** * 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* root) { return root ? isSymmetric(root->left,root->right) : true; } bool isSymmetric(TreeNode *left,TreeNode *right) { if(!left && !right) return true;//终止条件 if(!left || !right) return false;//终止条件 return left->val== right->val && isSymmetric(left->left,right->right) && isSymmetric(left->right,right->left);//三方合并 } };
版权声明:本文为博主原创文章,未经博主允许不得转载。
标签:
原文地址:http://blog.csdn.net/flyljg/article/details/48031249