[leetcode]31 Symmetric Tree

本文主要是介绍[leetcode]31 Symmetric Tree，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

题目链接：https://leetcode.com/problems/symmetric-tree/
Runtimes：8ms

1、问题

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.

2、分析

从左到右， 从右到左，两个操作同时进行，对比看看是否相等，不等则返回false。

3、小结

以为会花多时间，没想到跑得还挺快的。

4、实现

/*** 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 isSymmetric(TreeNode *root) {if(NULL == root)return true;vector<TreeNode *> v1, v2;v1.push_back(root);v2.push_back(root);while(v1.size() > 0 && v2.size() > 0){TreeNode *t1 = v1.back(); v1.pop_back();TreeNode *t2 = v2.back(); v2.pop_back();if(t1->val != t2->val)return false;if(NULL != t1->left && NULL != t2->right){v1.push_back(t1->left);v2.push_back(t2->right);}else if(NULL == t1->left && NULL == t2->right){}else{return false;}if(NULL != t1->right && NULL != t2->left){v2.push_back(t1->right);v1.push_back(t2->left);}else if(NULL == t1->right && NULL == t2->left){}else{return false;}}return true;}
};

5、反思

树的非递归操作，熟练一些，一气呵成。

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