101. Symmetric Tree（DFS）

1. Description

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree [1,2,2,3,4,4,3] is symmetric:

    1   / \  2   2 / \ / \3  4 4  3

But the following [1,2,2,null,3,null,3] is n

    1   / \  2   2   \   \   3    3

2. Analysis

直接的想法就是取出每一层，在该层做左右对称比较。思想与逐层遍历是类似的。时间复杂度为O(n)

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3. Algorithm achievement

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    bool isSymmetric(TreeNode* root) {    /*先把每一层分离出来，再前后对称比较*/    if(root == NULL) return true;    if(root->left == NULL && root->right == NULL) return true;    if(root->left == NULL ||  root->right == NULL) return false;    vector<TreeNode*> tmp;    TreeNode* node = NULL;    tmp.push_back(root);    while(!tmp.empty()) {        for(int i = 0, n = tmp.size(); i < n; i++) {            node = tmp.front();            tmp.erase(tmp.begin());            /*判断条件这里需要细心，有点坑*/            if(node != NULL)  {                tmp.push_back(node->left);                tmp.push_back(node->right);            }        }        if(tmp.size()%2 != 0) return false;        for(int i = 0, j = tmp.size()-1; i < tmp.size()/2; i++, j--) {            if(tmp[i] == NULL &&  tmp[j] == NULL ) continue;            if(tmp[i] == NULL || tmp[j] == NULL ) return false;            if(tmp[i] != NULL && tmp[j] != NULL) {                 if((tmp[i])->val != (tmp[j])->val)                     return false;            } else if(tmp[i] != NULL || tmp[j] != NULL)                 return false;        }    }    return true;    }    };

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贴个图防坑！注意NULL结点也会参与比较的，所以需要接入向量！

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