[leetcode]31 Symmetric Tree

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。