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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}".class Solution { public: bool isSymmetric(TreeNode *root){ return root? Symmetric(root->left,root->right):true; } bool Symmetric(TreeNode *left, TreeNode *right){ if(left==NULL && right==NULL) return true;// 左右孩子为空 if(!left || !right) return false; // 仅含有左子树或者右子树 return left->val == right->val && Symmetric(left->left,right->right) && Symmetric(left->right,right->left); } };
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[LeetCode 题解]: Symmetric Tree,布布扣,bubuko.com
标签：style blog http color strong io for ar
原文地址：http://www.cnblogs.com/double-win/p/3891215.html
