题目：

题解：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
bool isSymmetric(struct TreeNode* root) {

    if (root == NULL) return true;//如果根为空直接返回true
    
    struct TreeNode* p = root->left;
    struct TreeNode* q = root->right;//一个指向左子树，一个指向右子树

    struct TreeNode** stk1 = malloc(550 * sizeof(struct TreeNode));
    struct TreeNode** stk2 = malloc(550 * sizeof(struct TreeNode));//创建栈
    int top1 = 0, top2 = 0;//创建栈顶指针

    while ((p != NULL || q != NULL) || (top1 != 0 && top2 != 0)) {//栈不空或结点不全为空时循环

        if (p != NULL && q != NULL) {//两个结点都不空时

            if (p->val != q->val) return false;//结点值不同直接false
            
            stk1[top1++] = p;
            stk2[top2++] = q;//入栈
            p = p->left;
            q = q->right;//指针镜像移动
 
        }
        else if (p == NULL && q == NULL) {//两个结点都为空时

            p = stk1[--top1];
            q = stk2[--top2];//接收栈顶结点
            p = p->right;
            q = q->left;//指针镜像移动

        }
        else return false;//如果一个结点为空而另一个不为空时，直接false
        
    }

    return true;//经过以上处理后，此时栈已经空了，并且所有结点已经镜像遍历并比较过了，直接返回true

}