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\n");

        }else{

            printf("YES\n");

        }



    }

    return 0;

}

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