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