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 binary tree * 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 isMirror(root->left,root->right); } bool isMirror(TreeNode* left,TreeNode* right) { if(left==NULL && right==NULL) return true; if(left==NULL || right==NULL) return false; if(left->val!=right->val) return false; return isMirror(left->left,right->right) && isMirror(left->right,right->left); } };
原文：http://www.cnblogs.com/erictanghu/p/3759592.html