[刷题] PTA 03-树1 树的同构
程序：
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; //将根结点置为空，若为空树时返回Null 16 char cl,cr; 17 scanf("%d\n",&N); 18 if(N) { 19 for(i=0; i<N; i++) check[i]=0; //将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 //都为空树则同构 40 if((R1==Null)&&(R2==Null)) 41 return 1; 42 //只有一个根结点为空则不同构 43 if(((R1==Null)&&(R2!=Null))||((R1!=Null)&&(R2==Null))) 44 return 0; 45 //根结点数据不同则不同构 46 if(T1[R1].Element!=T2[R2].Element) 47 return 0; 48 //左儿子都为空，判断右儿子是否同构 49 if((T1[R1].Left==Null)&&(T2[R2].Left==Null)) 50 return Isomorphic(T1[R1].Right,T2[R2].Right); 51 //左儿子结点都不为空且数据相等，对左儿子的左右子树进行递归 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 //左儿子不一样，左右交换后递归 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、用数组存储树，结点无序
2、需找出树的根结点
3、判断是否同构要考虑周全
