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.
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}".
Solution 1: 递归，left对应right，left->left对应right->right，left->right对应right->left
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 bool isSymmetric(TreeNode* root) { 13 if(!root)return true; 14 return isSymTree(root->left,root->right); 15 } 16 bool isSymTree(TreeNode *p,TreeNode *q){ 17 if(!isSameNode(p,q)) 18 return false; 19 else if(!p&&!q) 20 return true; 21 else 22 return isSymTree(p->left,q->right) && isSymTree(p->right,q->left); 23 } 24 bool isSameNode(TreeNode *p,TreeNode *q){ 25 if(!p&&!q) //必需加上这个判断条件，否则若p、q为空下面的->val会运行时错误 26 return true; 27 else if((!p&&q)||(p&&!q)||(p->val!=q->val)) //利用||的短路效应避免运行时错误 28 return false; 29 return true; 30 } 31 };
Solution 2 :非递归，待续
【LeetCode】101 - Symmetric Tree
原文：http://www.cnblogs.com/irun/p/4719686.html