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.
OJ's Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.
Here's an example:
1 / 2 3 / 4 5
The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}"
.
/** * 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 check(TreeNode* left,TreeNode* right){ if(left==NULL&&right==NULL)return true; else if(left!=NULL&&right!=NULL){ if(left->val!=right->val)return false; else return check(left->left,right->right)&&check(left->right,right->left); } else return false; } bool isSymmetric(TreeNode *root) { // Start typing your C/C++ solution below // DO NOT write int main() function if(root==NULL)return true; return check(root->left,root->right); } };