题目描述

blablabla 加入了负数样本

样例

blablabla

算法1

(暴力枚举) $O(n^2)$

blablabla

时间复杂度分析：blablabla

C++ 代码

blablabla

算法2

(暴力枚举) $O(n^2)$

blablabla

时间复杂度分析：blablabla

C++ 代码

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {//是不是因为没有加入负数判断？？对！yxc的做法没有加入负数判断！
public:

    // Encodes a tree to a single string.
    string serialize(TreeNode* root) {
        string res;
        dfs_s(root, res);
        return res;
    }

    void dfs_s(TreeNode *root, string &res){
        if(!root){
            res += "null ";
            return;
        }
        res += to_string(root->val) + ' ';
        dfs_s(root->left, res);
        dfs_s(root->right, res);
    }

    // Decodes your encoded data to tree.
    TreeNode* deserialize(string data) {
        int u = 0;
        return dfs_d(data, u);
    }

    TreeNode* dfs_d(string data, int &u){
        if(u == data.size()) return nullptr;
        int k = u;
        while (data[k] != ' ') k ++ ;
        if (data[u] == 'n'){
            u = k + 1;
            return nullptr;
        }

        int val = 0;
        bool is_neg = false;
        if(data[u] == '-') is_neg = true, u ++;
        for(int i = u; i < k; i ++ ) val = val * 10 + data[i] - '0';
        if(is_neg) val = - val;
        u = k + 1;
        auto root = new TreeNode(val);
        root->left = dfs_d(data, u);
        root->right = dfs_d(data, u);
        return root;
    }
};