AcWing 50. 序列化二叉树
宋标 Lv5
  2022-08-18 02:31:01 2022-08-18 02:31    

题目

请实现两个函数，分别用来序列化和反序列化二叉树。

您需要确保二叉树可以序列化为字符串，并且可以将此字符串反序列化为原始树结构。

数据范围

树中节点数量 。

样例

你可以序列化如下的二叉树
    8
   / \
  12  2
     / \
    6   4

为："[8, 12, 2, null, null, 6, 4, null, null, null, null]"

注意:

  • 以上的格式是AcWing序列化二叉树的方式，你不必一定按照此格式，所以可以设计出一些新的构造方式。

题解

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

#define N 1010

using namespace std;

class Solution {
public:

    string tree_seri;
    int pr_or[N], in_or[N];
    int cnt;

    void set(int num)
    {
        if (!num) return;
        set(num / 10);
        if (num < 0) num *= -1;
        tree_seri += num % 10 + '0';
    }

    void order(TreeNode* root, bool pre)
    {
        if (!root) return;

        if (!pre) order(root -> left, pre); 

        int &v = root -> val;
        if (!v) tree_seri += '0';
        else 
        {
            if (v < 0) tree_seri += '-';
            set(v);
        }
        tree_seri += ",";

        if (pre) order(root -> left, pre); 

        order(root -> right, pre);
    }

    // Encodes a tree to a single string.
    string serialize(TreeNode* root) {

        if (!root) return "";

        order(root, true);
        tree_seri[tree_seri.size() - 1] = '|';
        order(root, false);
        tree_seri[tree_seri.size() - 1] = '\0';

        return tree_seri;
    }



    void arr_set(int begin, int end, int arr[])
    {
        bool neg = false;
        for (int i = begin; i < end; ++ i)
        {
            if (tree_seri[i] == ',') 
                continue;
            if (tree_seri[i] == '-') 
            {
                neg = true;
                continue;
            }

            int sum = 0;
            while (tree_seri[i] >= '0' && tree_seri[i] <= '9') 
            {
                sum += tree_seri[i] - '0';
                sum *= 10;
                ++ i;
            }
            -- i;

            if (neg) sum *= -1, neg = false;
            arr[cnt ++] = sum / 10;
        }
    }

    map<int,int> hash;

    TreeNode* dfs(int pl, int pr, int il, int ir)
    {
        if (pl > pr) return nullptr;
        auto *root = new TreeNode(pr_or[pl]);
        int k = hash[pr_or[pl]];
        root -> left = dfs(pl + 1, pl + k - il, il, k - 1);
        root -> right = dfs(pl + k - il + 1, pr, k + 1, ir);
        return root;
    }

    // Decodes your encoded data to tree.
    TreeNode* deserialize(string data) {

        if (data == "") return nullptr;

        tree_seri = data;

        int begin = 0, end = 0;
        while (data[end] != '|') ++ end;

        arr_set(begin, end, pr_or); cnt = 0;
        arr_set(end + 1, data.size() - 1, in_or);

        for (int i = 0; i < cnt; ++ i) hash[in_or[i]] = i;
        return dfs(0, cnt - 1, 0, cnt - 1);
    }
};
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