给定一个二叉树，检查它是否是镜像对称的。

例如，二叉树 [1,2,2,3,4,4,3] 是对称的。

1

/
2 2
/ \ /
3 4 4 3

但是下面这个 [1,2,2,null,3,null,3] 则不是镜像对称的:

1

/
2 2
\
3 3

---

方法1

根左右遍历一次树得到数组，根右左遍历一次得到数组
对比，相等则对称
存在根左右与根右左相等的子树，便使用某个数来标记全部空结点防止出现这种情况

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    void leorder(TreeNode*root,vector <int>&arrleft)
    {
        if(!root){arrleft.push_back(-1);return;}
        arrleft.push_back(root->val);
        leorder(root->left,arrleft);
        leorder(root->right,arrleft);

    }
    void riorder(TreeNode*root,vector <int>&arrright)
    {
        if(!root){arrright.push_back(-1);return;}
        arrright.push_back(root->val);
        riorder(root->right,arrright);
        riorder(root->left,arrright);

    }
    bool isSymmetric(TreeNode* root) {
        vector<int>arrleft;
        vector<int>arrright;
        leorder(root,arrleft);
        riorder(root,arrright);
        if(arrleft==arrright)return 1;
        else return 0;


    }
};

当然，这个方法不够聪明

---

方法二：

使用两个指针分部从左边右边遍历即可，使用原理为递归

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    bool check(TreeNode*p,TreeNode* q)
    {
        if(!q&&!p)return true;
        if(!p||!q)return false;
        return (p->val==q->val)
        &&check(p->left,q->right)
        &&check(p->right,q->left);
    }

    bool isSymmetric(TreeNode* root) {
        return check(root,root);
    }
};

方法三：

利用队列，类似于层序遍历来迭代它，也可以说是广度搜索（或许

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    bool check(TreeNode* u,TreeNode* v)
    {
        queue<TreeNode*>que;
        que.push(u);que.push(v);
        while(!que.empty())
        {  u=que.front();que.pop();
           v=que.front();que.pop();
        if(!u&&!v)continue;
        if((!u||!v)||u->val!=v->val)return false;
        que.push(u->left);
        que.push(v->right);

        que.push(u->right);
        que.push(v->left);

        }
        return true;
    }
    bool isSymmetric(TreeNode* root) {
        return check(root,root);

    }
};