如何实现创建固定结点个数的二叉树？

Question

如图所示，二叉树结点为7个，如果用递归或者基于栈的方式，创建函数应该怎么去编写呢？

Answer 1

#include<stdio.h>
#define max 7
#define OK 1
typedef struct SqBiTree
{
    char node[max+1];
}
SqBiTree;
//SqBiTree T;
int CreatBitree(SqBiTree *T)
{
    T->node[0]=-1;
    printf("输入二叉树：");
    for(int i=1; i<=max; i++)
    {
        scanf("%c",&T->node[i]);
    }
    //put(T);
    return OK;
}
void put(SqBiTree T)
{
    for(int i=1; i<=max; i++)
        printf("%c ",T.node[i]);
    printf("\n");
}
int Nonleafnode(SqBiTree T)
{
    int count=0;
    for(int i=1; i<=max; i++)
    {
        if(T.node[i]!='.')
        {
            if(i*2<=max||i*2+1<=max)
            {
                count++;
                //printf("%c\n",T.node[i]);
            }
        }
    }
    return count;
}
void Nonemptynodewithoutparents(SqBiTree T)
{
    for(int i=1; i<=max; i++)
    {
        if(T.node[i]!='.'&&(T.node[i/2]=='.'&&T.node[(i-1)/2]=='.'))
        {
            printf("出现无双亲的非空节点%c\n",T.node[i]);
        }
    }
}
int main()
{
    SqBiTree T;
    int oper=1;
    while(oper!=-1)
    {
        printf("1.建立二叉树\n");
        printf("2.打印无双亲的非空节点\n");
        printf("3.打印非叶子节点数目\n");
        scanf("%d",&oper);
        getchar();
        switch (oper)
        {
        case 1:
            CreatBitree(&T);
            break;
        case 2:
            Nonemptynodewithoutparents(T);
            break;
        case 3:
            printf("非叶子节点数目：%d\n",Nonleafnode(T));
            break;
        }
    }
    // CreatBitree(&T);
    //Nonemptynodewithoutparents(T);
    //put(T);
    return 0;
}