面试题37：序列化二叉树

输入一棵二叉树，将其序列化为char*，并且支持反序列化为二叉树。

解题思路

• 先序遍历二叉树，序列化
• 同先序遍历结构，反序列化

代码运行结果没问题，但是未通过所有测试案例。。。

#include <iostream>
#include <algorithm>
#include <math.h>
#include <cstring>
#include "ListNode.h"
#include "TreeNode.h"
#include "Graph.h"
using namespace std;

#define MAXNUM 100010
#define DRIFT 1001

void Serialize(TreeNode* pRoot, vector<char> &vec){
    if(pRoot == nullptr){
        vec.push_back('$');
        vec.push_back(',');
        return ;
    }
    vec.push_back('0' + pRoot->val);
    vec.push_back(',');
    Serialize(pRoot->left, vec);
    Serialize(pRoot->right, vec);
}

// 序列化二叉树（前序遍历的格式）
char* Serialize(TreeNode* root){
    vector<char> vecStr;

    Serialize(root, vecStr);
    char* str = new char[vecStr.size() + 1];
    int i;
    for(i = 0; i < (int)vecStr.size(); i++)
        str[i] = vecStr[i];

    str[i] = '';
    return str;
}

TreeNode* Deserialize(vector<char> &vec){
    if(vec.size() == 0)
        return nullptr;

    TreeNode* pNode = new TreeNode(vec[vec.size() - 1] - '0');
    vec.pop_back();
    vec.pop_back();

    // 如果读取的是数字
    if(vec[vec.size() - 1] >= '0' && vec[vec.size() - 1] <= '9')
        pNode->left = Deserialize(vec);
    else{
        pNode->left = nullptr;
        vec.pop_back();
        vec.pop_back();
    }

    // 如果读取的是数字
    if(vec[vec.size() - 1] >= '0' && vec[vec.size() - 1] <= '9')
        pNode->right = Deserialize(vec);
    else{
        pNode->right = nullptr;
        vec.pop_back();
        vec.pop_back();
    }
    return pNode;
}

// 根据前序遍历的序列化结果，重建二叉树
TreeNode* Deserialize(char *str){

    // 根据前序遍历重建
    vector<char> vec;
    string s = str;
    for(int i = s.length() - 1, j = 0; i >= 0; i--)
        vec.push_back(s[i]);

    return Deserialize(vec);
}

int main()
{
    TreeNode* pNode1 = CreateBinaryTreeNode(1);
    TreeNode* pNode2 = CreateBinaryTreeNode(2);
    TreeNode* pNode3 = CreateBinaryTreeNode(3);
    TreeNode* pNode4 = CreateBinaryTreeNode(4);
    TreeNode* pNode5 = CreateBinaryTreeNode(5);
    TreeNode* pNode6 = CreateBinaryTreeNode(6);

    ConnectTreeNodes(pNode2, pNode4, nullptr);
    ConnectTreeNodes(pNode3, pNode5, pNode6);
    ConnectTreeNodes(pNode1, pNode2, pNode3);

    // 如果写为char str[100]返回的话，会报错，不能返回局部变量
    char* str = Serialize(pNode1);
    cout<<str<<endl;

    PrintTree(Deserialize(str));
    delete str;
    return 0;
}