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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.
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}".
Tree Depth-first Search
/** * 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 dfscheck(TreeNode *leftNode,TreeNode *rightNode){ if(leftNode==NULL &&rightNode==NULL) return true; if(leftNode==NULL || rightNode==NULL) return false; return leftNode->val==rightNode->val && dfscheck(leftNode->left,rightNode->right) && dfscheck(leftNode->right,rightNode->left); } bool isSymmetric(TreeNode *root) { if(root==NULL) return true; return dfscheck(root->left,root->right); } };
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原文地址：http://www.cnblogs.com/li303491/p/4108903.html
