题目1520：树的子结构 ——树的遍历

http://ac.jobdu.com/problem.php?pid=1520

题目描述：

输入两颗二叉树A，B，判断B是不是A的子结构。

先建树，为了后面匹配时提高速度，每个结点做一个索引

然后枚举A中的结点是否与B的树根相同，若相同，则遍历B的同时遍历A,判断是否相似

#include<stdio.h>

struct Tree{
    int v;
    Tree *left,*right;
}* Atree_p[1099],* Btree_p[1099];

int n,m;
int nodeL,nodeR;
void build(Tree *thead,int num,int v){
    thead->v=v;
    if(num==0){
        thead->left=NULL;
        thead->right=NULL;
    }
    if(num==1){
        Atree_p[nodeL] =thead->left=new Tree;
        thead->right=NULL;
    }
    if(num==2){
        Atree_p[nodeL] = thead->left=new Tree;
        Atree_p[nodeR] = thead->right=new Tree;
    }
}

void build_B(Tree *thead,int num,int v){
    thead->v=v;
    if(num==0){
        thead->left=NULL;
        thead->right=NULL;
    }
    if(num==1){
        Btree_p[nodeL] =thead->left=new Tree;
        thead->right=NULL;
    }
    if(num==2){
        Btree_p[nodeL] = thead->left=new Tree;
        Btree_p[nodeR] = thead->right=new Tree;
    }
}

int AtreeV[1099],BtreeV[1099];

int dif=0;
int dfs(Tree* Athead,Tree *Bthead){
    if(dif==1)return 1;
    if(Bthead->left!=NULL){
        if(Athead->left==NULL||(Athead->left->v)!=(Bthead->left->v)){
            dif=1;
            return 1;
        }
        dfs(Athead->left,Bthead->left);
    }
        
    if(Bthead->right!=NULL){
        if(Athead->right==NULL||(Athead->right->v)!=(Bthead->right->v)){
            dif=1;
            return 1;
        }
        dfs(Athead->right,Bthead->right);
    }
    if(dif==1)return 1;
    else return 0;
}

int find(){
    int i,j;
    for(i=1;i<=n;i++){
        if((Atree_p[i]->v)!=(Btree_p[1]->v))continue;
        dif=0;
        if(dfs(Atree_p[i],Btree_p[1])==0) return 1;
    }
    return 0;
}

int main(){
    int i,j,k,num,no;
    while(scanf("%d%d",&n,&m)!=EOF){
        for(i=1;i<=n;i++){
            scanf("%d",&AtreeV[i]);
        }
        Atree_p[1]=new Tree;
        for(i=1;i<=n;i++){
            scanf("%d",&num);
            if(num==1){
                scanf("%d",&nodeL);
            }
            if(num==2){
                scanf("%d%d",&nodeL,&nodeR);
            }build(Atree_p[i],num,AtreeV[i]);
        }
        
        //B
        for(i=1;i<=m;i++){
            scanf("%d",&BtreeV[i]);
        }
        Btree_p[1]=new Tree;
        for(i=1;i<=m;i++){
            scanf("%d",&num);
            if(num==1){
                scanf("%d",&nodeL);
            }
            if(num==2){
                scanf("%d%d",&nodeL,&nodeR);
            }build_B(Btree_p[i],num,BtreeV[i]);
        }
        //B为空树时不是任何树的子树
        if(m==0||find()==0){
            printf("NO
");
        }else{
            printf("YES
");
        }

    }
    return 0;
}