03-树1 树的同构

　　这一题是课上例题，用到的递归思想较为关键。

#include <stdio.h>
#include <stdlib.h>
#define MaxTreeNodes 10

struct TreeNode{
    char Element;
    int Left;
    int Right;
};
typedef struct TreeNode FullTree;
FullTree T1[MaxTreeNodes], T2[MaxTreeNodes];

int BuildTree(FullTree T[])
{
    int Root;
    int i = -1, N;
    int check[10];
    char cl, cr;
    scanf("%d
", &N);  //'
'一定要加！！
    if(N){
        for(i = 0; i < N; i++){
            check[i] = 0;  //check[N]中没有被指向都为0，指向了就赋值为1，最后没被指向的就是根节点
        }
        for(i = 0; i < N; i++){
            scanf("%c %c %c
", &T[i].Element, &cl, &cr);
            if(cl == '-')
                T[i].Left = -1;
            else{
                T[i].Left = cl - '0';
                check[T[i].Left] = 1;
            }
            if(cr == '-')
                T[i].Right = -1;
            else{
                T[i].Right = cr - '0';
                check[T[i].Right] = 1;
            }
        }
        for(i = 0; i < N; i++){
            if(!check[i])
                break;
        }
    }
    Root = i;
    return Root;
}

int isOmorphic(int R1, int R2)
{
    if((R1 == -1) && (R2 == -1))
        return 1;
    if(((R1 != -1) && (R2 == -1)) || ((R1 == -1) && (R2 != -1)))
        return 0;
    if(T1[R1].Element != T2[R2].Element)
        return 0;
    if((T1[R1].Left == -1) && (T2[R2].Left == -1))
        return isOmorphic(T1[R1].Right, T2[R2].Right);
    if((T1[R1].Left != -1) && (T2[R2].Left != -1)
        && (T1[T1[R1].Left].Element) == (T2[T2[R2].Left].Element))
        return (isOmorphic(T1[R1].Left, T2[R2].Left) && isOmorphic(T1[R1].Right, T2[R2].Right));
    else
        return (isOmorphic(T1[R1].Left, T2[R2].Right) && isOmorphic(T1[R1].Left, T2[R2].Right));
}

int main(int argc, char const *argv[])
{
    int R1, R2;
    R1 = BuildTree(T1);
    R2 = BuildTree(T2);
    if(isOmorphic(R1, R2))
        printf("Yes
");
    else
        printf("No
");
    return 0;
}

　　一开始没有加第20行的if(N)导致空树情况下总是错误，后来意识到虽然空树时没有进入循环，但是循环判断条件if语句还是执行了的，期间会使得i重置为0，使得Root结果错误。