101题

方法一：使用广度遍历（BFS），通过输入两个根节点将对称的两个结对送入到queue中，并且两两释放
遇到问题：continue的使用方法要注意

/**
 * 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 isSymmetric(TreeNode* root) {
        bool ret=check(root,root);
        return ret;
    }
    bool check(TreeNode* u,TreeNode* v){
        if(u==NULL&&v==NULL){
            return true;
        }
        else if(u==NULL||v==NULL){
            return false;
        }
        queue<TreeNode*>q;
        q.push(u);
        q.push(v);
        while(!q.empty()){
            TreeNode* head1=q.front();
            q.pop();
            TreeNode* head2=q.front();
            q.pop();
            if(head1==NULL&&head2==NULL){
            continue;
            }
            else if(head1==NULL||head2==NULL){
            return false;
            }
            if(head1->val!=head2->val){
                return false;
            }
            q.push(head1->left);
            q.push(head2->right);
            q.push(head1->right);
            q.push(head2->left);
        }
        return true;
    }
};

方法二：递归

class Solution {
public:
    bool isSymmetric(TreeNode* root) {
        bool ret;
        ret=check(root,root);
        return ret;
    }
    bool check(TreeNode* u,TreeNode* v){
        if(u==NULL&&v==NULL){
            return ret;
        }
        else if(u==NULL||v==NULL){
            return ret= false;
        }
        if(u->val!=v->val){
            ret= false;
        }
        ret=check(u->left,v->right);
        ret=check(u->right,v->left);
        return ret;
    }
private:
    bool ret=true;
};