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.
/** * 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) { if(!root) return true; else return isSymmetric(root->left,root->right); } bool isSymmetric(TreeNode *left,TreeNode *right) { if(!left&&!right) return true; if(!left||!right) return false; return left->val==right->val&&isSymmetric(left->left,right->right)&&isSymmetric(left->right,right->left); } };