解读赫夫曼树编码的问题

#include<iostream>
#include<assert.h>
using namespace std;

struct HuffmanNode
{
    unsigned int weight;
    unsigned int parent,leftChild,rightChild;
    HuffmanNode()
    {
        weight=0;parent=0;leftChild=0;rightChild=0;
    }
};

void Select(const HuffmanNode* & nodelist,const int length,int & a, int &b)
{
    int min=1000000,min2=1000000;
    for(int i=0;i<length;i++)
    {
        if(min>nodelist[i].weight&&nodelist[i].parent==0)
        {
            min=nodelist[i].weight;
            a=i;
        }
    }
    for(int j=0;j<length;j++)
    {
        if(j!=a&&min2>nodelist[j].weight&&nodelist[j].parent==0)
        {
            min2=nodelist[j].weight;
            b=j;
        }
    }
}

char ** HuffmanCoding(const int *w, const int n)
{
    assert(w!=NULL);
    int i,min1,min2;
    int m = 2*n-1;
    HuffmanNode * nodelist = new HuffmanNode[m]();
    for(i=0;i<n;i++)
    {
        nodelist[i].weight=w[i];
        nodelist[i].parent=0;
    }
    for(i=n;i<m;i++)
    {
        Select(nodelist,i,min1,min2);
        nodelist[min1].parent=i;
        nodelist[min2].parent=i;
        nodelist[i].weight=nodelist[min1].weight+nodelist[min2].weight;
        nodelist[i].rightChild=min2;
        nodelist[i].leftChild=min1;
        nodelist[i].parent=0;
    }
    char temp [20];
    char ** code = new char * [n];
    for(i=0;i<n;i++)
    {
        int j=i;
        int index=0;
        while(j!=m-1)
        {
            if(j==nodelist[nodelist[j].parent].leftChild) temp[index++]='0';
            else temp[index++]='1';
            j=nodelist[j].parent;
        }
        temp[index]='\0';
        code[i] = new char[index+1];
        strcpy(code[i],temp);
    }
    delete nodelist;
    return code;
}

int main()
{
    const int size=6;
    char word[size]={'A','B','C','D','E','F'};//编码字符
    int w[size]={4,3,2,1,7,8};//权重
    char ** code;
    code=HuffmanCoding(w,size);
    assert(code!=NULL);
    for(int i=0;i<size;i++)
    {
        cout<<word[i]<<" is coded as "<<code[i]<<endl;
    }
    //注意二级指针的释放问题
    for(int j=0;j<size;j++)
    {
        delete []code[j];
    }
    delete []code;
    return 0;
}