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。

 1 #include <stdio.h>
 2 #include <stdlib.h>
 3 #define MaxTreeNodes 10
 4 
 5 struct TreeNode{
 6     char Element;
 7     int Left;
 8     int Right;
 9 };
10 typedef struct TreeNode FullTree;
11 FullTree T1[MaxTreeNodes], T2[MaxTreeNodes];
12 
13 int BuildTree(FullTree T[])
14 {
15     int Root;
16     int i = -1, N;
17     int check[10];
18     char cl, cr;
19     scanf("%d\n", &N);  //'\n'一定要加！！
20     if(N){
21         for(i = 0; i < N; i++){
22             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
23         }
24         for(i = 0; i < N; i++){
25             scanf("%c %c %c\n", &T[i].Element, &cl, &cr);
26             if(cl == '-')
27                 T[i].Left = -1;
28             else{
29                 T[i].Left = cl - '0';
30                 check[T[i].Left] = 1;
31             }
32             if(cr == '-')
33                 T[i].Right = -1;
34             else{
35                 T[i].Right = cr - '0';
36                 check[T[i].Right] = 1;
37             }
38         }
39         for(i = 0; i < N; i++){
40             if(!check[i])
41                 break;
42         }
43     }
44     Root = i;
45     return Root;
46 }
47 
48 int isOmorphic(int R1, int R2)
49 {
50     if((R1 == -1) && (R2 == -1))
51         return 1;
52     if(((R1 != -1) && (R2 == -1)) || ((R1 == -1) && (R2 != -1)))
53         return 0;
54     if(T1[R1].Element != T2[R2].Element)
55         return 0;
56     if((T1[R1].Left == -1) && (T2[R2].Left == -1))
57         return isOmorphic(T1[R1].Right, T2[R2].Right);
58     if((T1[R1].Left != -1) && (T2[R2].Left != -1)
59         && (T1[T1[R1].Left].Element) == (T2[T2[R2].Left].Element))
60         return (isOmorphic(T1[R1].Left, T2[R2].Left) && isOmorphic(T1[R1].Right, T2[R2].Right));
61     else
62         return (isOmorphic(T1[R1].Left, T2[R2].Right) && isOmorphic(T1[R1].Left, T2[R2].Right));
63 }
64 
65 int main(int argc, char const *argv[])
66 {
67     int R1, R2;
68     R1 = BuildTree(T1);
69     R2 = BuildTree(T2);
70     if(isOmorphic(R1, R2))
71         printf("Yes\n");
72     else
73         printf("No\n");
74     return 0;
75 }

　　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。

posted @ 2018-03-26 17:30  BianK  阅读(185)  评论(0)  编辑  收藏  举报