Symmetric Tree

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

进行两次深度优先搜索，分别对左子树和右子树优先，将遍历的结果保存起来，最后比较遍历结果即可。需要注意的是，当某子树为空的时候需要用0来标识，如树

    1   / \  2     

不加0的话 遍历结构都是（1，2），标识之后结构为（1,2,0）和（1,0,2）可判断出非对称树。

/** * 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) {        vector<int > v1,v2;        dfs1(root,v1);        dfs2(root,v2);        if(v1==v2)             return true;        else return false;    }private:    void dfs1(TreeNode *root,vector<int> & v)    {        if(root==NULL)        {            v.push_back(0);            return ;        }        v.push_back(root->val);        dfs1(root->left,v);        dfs1(root->right,v);    }    void  dfs2(TreeNode *root,vector<int> & v)    {        if(root==NULL)        {            v.push_back(0);            return ;        }        v.push_back(root->val);        dfs2(root->right,v);        dfs2(root->left,v);    }    };