[刷题] PTA 03-树1 树的同构

程序：

#include <stdio.h>
#define MaxTree 10
#define ElementType char
#define Tree int
#define Null -1

struct TreeNode {
    ElementType Element;
    Tree Left;
    Tree Right;
} T1[MaxTree],T2[MaxTree];
int N,check[MaxTree];

Tree BuildTree(struct TreeNode T[]) {
    int Root=Null,i;  //将根结点置为空，若为空树时返回Null 
    char cl,cr;
    scanf("%d
",&N);
    if(N) {
        for(i=0; i<N; i++) check[i]=0;  //将check[]置0 
        for(i=0; i<N; i++) {
            scanf("%c %c %c
",&T[i].Element,&cl,&cr);
            if(cl!='-') {
                T[i].Left=cl-'0';
                check[T[i].Left]=1;
            } else T[i].Left=Null;
            if(cr!='-') {
                T[i].Right=cr-'0';
                check[T[i].Right]=1;
            } else T[i].Right=Null;
        }
        for(i=0; i<N; i++)
            if(!check[i]) break;
        Root=i;
    }
    return Root;
}

int Isomorphic(Tree R1,Tree R2) {
    //都为空树则同构
    if((R1==Null)&&(R2==Null))   
        return 1;
    //只有一个根结点为空则不同构 
    if(((R1==Null)&&(R2!=Null))||((R1!=Null)&&(R2==Null))) 
        return 0;
    //根结点数据不同则不同构 
    if(T1[R1].Element!=T2[R2].Element)  
        return 0;
    //左儿子都为空，判断右儿子是否同构 
    if((T1[R1].Left==Null)&&(T2[R2].Left==Null))
        return Isomorphic(T1[R1].Right,T2[R2].Right);
    //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 
    if(((T1[R1].Left!=Null)&&(T2[R2].Left!=Null))&&
            ((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));
    //左儿子不一样，左右交换后递归 
    else
        return(Isomorphic(T1[R1].Left,T2[R2].Right)&&
               Isomorphic(T1[R1].Right,T2[R2].Left));
}

int main() {
    Tree R1,R2;
    R1=BuildTree(T1);
    R2=BuildTree(T2);
    if(Isomorphic(R1,R2))
        printf("Yes
");
    else printf("No
");
    return 0;
}

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