题目描述

序列化和反序列化二叉树

样例

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算法1

时间复杂度

参考文献

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 {
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)//加引用！！！！
    {
        if (u == data.size()) return NULL;
        int k = u;
        while (data[k] != ' ') k ++;
        if (data[u] == 'n') {
            u = k + 1;
            return NULL;
        }
        int val = 0;
        //负的
        if(data[u] == '-'){
            for (int i = u+1; i < k; i++) val = val * 10 + data[i] - '0';
                val  = -val;
        }
        else{
        //如果是数字是正的
        for (int i = u; i < k; i++) val = val * 10 + data[i] - '0';
        }
        u = k + 1;
        auto root = new TreeNode(val);
        root->left = dfs_d(data, u); //前序遍历的顺序取的
        root->right = dfs_d(data, u);
        return root;
    }
};

算法2

(暴力枚举) $O(n^2)$

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时间复杂度

参考文献

C++ 代码

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