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?
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ struct TreeNode { int val; TreeNode *left; TreeNode *right; TreeNode(int x) : val(x), left(NULL), right(NULL) {} }; class Solution { public: bool isSymmetric(TreeNode *root) { if(root == NULL) return true; return dfs(root->left,root->right); } private: bool dfs(TreeNode *a,TreeNode *b) { if(a == NULL && b == NULL) return true; if((a == NULL && b != NULL) || (a != NULL && b == NULL)) return false; return a->val == b->val && dfs(a->left,b->right) && dfs(a->right,b->left); } };
转载于:https://www.cnblogs.com/bhlsheji/p/5171069.html
相关资源：数据结构—成绩单生成器