题解 | #对称的二叉树# deque双向队列

deque支持随机读取，同时又有queue的特性，所以只要使用一个即可

/*
struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
    TreeNode(int x) :
            val(x), left(NULL), right(NULL) {
    }
};
*/
class Solution {
public:
    bool isSymmetrical(TreeNode* pRoot) {
        if(!pRoot) return true;
        deque<TreeNode*> d;
        d.push_back(pRoot);
        while(!d.empty()){
            int n = d.size();
            for(int i = 0; i<n; i++){
                TreeNode* cur = d.front();
                d.pop_front();
                if(!cur) continue;
                if(cur->left) d.push_back(cur->left);
                else d.push_back(nullptr);
                if(cur->right) d.push_back(cur->right);
                else d.push_back(nullptr);
            }
            n = d.size();
            if(n%2) return false;
            for(int i = 0; i<n/2; i++){
                if(!d[i] && !d[n-1-i]) continue;
                else if(!d[i] || !d[n-1-i] || d[i]->val != d[n-1-i]->val) return false;
                
            }
        }
        return true;
    }

};