题解 | #牛群的轴对称结构#

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
#include <vector>
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param root TreeNode类 
     * @return bool布尔型
     */
    void leftPreOrder(TreeNode* root, vector<int> &result){
        if(root == nullptr){
            result.push_back(-1);
            return;
        }

        result.push_back(root->val);
        leftPreOrder(root->left, result);
        leftPreOrder(root->right, result);
    }

    void rightPreOrder(TreeNode* root, vector<int> &result){
        if(root == nullptr){
            result.push_back(-1);
            return;
        }

        result.push_back(root->val);
        rightPreOrder(root->right, result);
        rightPreOrder(root->left, result);
    }

    bool isSymmetric(TreeNode* root) {
        // write code here
        if (root == nullptr) {
            return true;
        }
        vector<int> left;
        vector<int> right;

        leftPreOrder(root->left, left);
        rightPreOrder(root->right, right);

        return left == right;
    }
};