/* 此树为

              A

        B         C        D

E        F     G

孩子-兄弟链表为

                        A

                B        

        E           C

           F     G        D

*/

本题代码如下

void createtree(tree* t, char a[], int degree[], int n)
{
    // 为B数组分配内存
    tree* B = (tree*)malloc(sizeof(tree) * n);
    int i = 0;
    int j = 0;
    int k = 0;
    int d = 0;
    for (i = 0; i < n; i++) {
        B[i] = (tree)malloc(sizeof(treenode)); // 为每个节点分配内存
        B[i]->data = a[i];
        B[i]->lchild = B[i]->rborther = NULL;
    }
    for (i = 0; i < n; i++) {
        d = degree[i];
        if (d) {
            k++;
            B[i]->lchild = B[k];
            for (j = 2; j <= d; j++) {
                k++;
                B[k - 1]->rborther = B[k];
            }
        }
    }
    *t = B[0]; // 返回根节点
}

完整测试代码为

#include<stdio.h>
#include<stdlib.h>
typedef struct treenode
{
    char data;
    struct treenode* lchild, * rborther;
} treenode, * tree;
void createtree(tree* t, char a[], int degree[], int n)
{
    // 为B数组分配内存
    tree* B = (tree*)malloc(sizeof(tree) * n);
    int i = 0;
    int j = 0;
    int k = 0;
    int d = 0;
    for (i = 0; i < n; i++) {
        B[i] = (tree)malloc(sizeof(treenode)); // 为每个节点分配内存
        B[i]->data = a[i];
        B[i]->lchild = B[i]->rborther = NULL;
    }
    for (i = 0; i < n; i++) {
        d = degree[i];
        if (d) {
            k++;
            B[i]->lchild = B[k];
            for (j = 2; j <= d; j++) {
                k++;
                B[k - 1]->rborther = B[k];
            }
        }
    }
    *t = B[0]; // 返回根节点
}
void print(tree* t)//先序输出树结点
{
    if (*t) {
        printf("%c ", (*t)->data); 
        print(&(*t)->lchild);
        print(&(*t)->rborther);
    }
}
int main()
{
    tree t;
    char a[7] = "ABCDEFG";
    int degree[7] = { 3,2,1,0,0,0,0 };
    createtree(&t, a, degree, 7);
    print(&t);
    return 0;
}