PTA

03-樹1 樹的同構

 1 #include <stdio.h>
 2 #define MaxTree 10
 3 #define ElementType char
 4 #define Tree int
 5 #define Null -1
 6 
 7 struct TreeNode {
 8     ElementType Element;
 9     Tree Left;
10     Tree Right;
11 } T1[MaxTree],T2[MaxTree];
12 int N,check[MaxTree];
13 
14 Tree BuildTree(struct
 
 TreeNode T[]) {
15     int Root=Null,i;  //將根結點置為空，若為空樹時返回Null 
16     char cl,cr;
17     scanf("%d\n",&N);
18     if(N) {
19         for(i=0; i<N; i++) check[i]=0;  //將check[]置0 
20         for(i=0; i<N; i++) {
21             scanf("%c %c %c\n",&T[i].Element,&cl,&cr);
22             if
 
(cl!=‘-‘) {
23                 T[i].Left=cl-‘0‘;
24                 check[T[i].Left]=1;
25             } else T[i].Left=Null;
26             if(cr!=‘-‘) {
27                 T[i].Right=cr-‘0‘;
28                 check[T[i].Right]=1;
29             } else T[i].Right=Null;
30         }
31         for
 
(i=0; i<N; i++)
32             if(!check[i]) break;
33         Root=i;
34     }
35     return Root;
36 }
37 
38 int Isomorphic(Tree R1,Tree R2) {
39     //都為空樹則同構
40     if((R1==Null)&&(R2==Null))   
41         return 1;
42     //只有一個根結點為空則不同構 
43     if(((R1==Null)&&(R2!=Null))||((R1!=Null)&&(R2==Null))) 
44         return 0;
45     //根結點數據不同則不同構 
46     if(T1[R1].Element!=T2[R2].Element)  
47         return 0;
48     //左兒子都為空，判斷右兒子是否同構 
49     if((T1[R1].Left==Null)&&(T2[R2].Left==Null))
50         return Isomorphic(T1[R1].Right,T2[R2].Right);
51     //左兒子結點都不為空且數據相等，對左兒子的左右子樹進行遞歸 
52     if(((T1[R1].Left!=Null)&&(T2[R2].Left!=Null))&&
53             ((T1[T1[R1].Left].Element)==(T2[T2[R2].Left].Element)))
54         return(Isomorphic(T1[R1].Left,T2[R2].Left)&&
55                Isomorphic(T1[R1].Right,T2[R2].Right));
56     //左兒子不一樣，左右交換後遞歸 
57     else
58         return(Isomorphic(T1[R1].Left,T2[R2].Right)&&
59                Isomorphic(T1[R1].Right,T2[R2].Left));
60 }
61 
62 int main() {
63     Tree R1,R2;
64     R1=BuildTree(T1);
65     R2=BuildTree(T2);
66     if(Isomorphic(R1,R2))
67         printf("Yes\n");
68     else printf("No\n");
69     return 0;
70 }

分析：

1、用數組存儲樹，結點無序

2、需找出樹的根結點

3、判斷是否同構要考慮周全

PTA——樹的同構