5-4 是否同一棵二叉搜索树

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先建一颗模板数，然后对于每一个测试，检测到达该节点的路径上的每一个点是否遍历过，如果遍历过，则相同，否则树的顺序就不一样

#include<stdio.h>#include<stdlib.h>#include<iostream>#include<queue>using namespace std;typedef struct TreeNode *Tree;struct TreeNode{    int v;    Tree right,left;    int flag;};Tree Insert(Tree T,int V){    if(!T)    {        T=(Tree)malloc(sizeof(struct TreeNode));        T->v=V;        T->left=T->right=NULL;        T->flag=0;    }    else    {        if(V > T->v)            T->right= Insert(T->right,V);        else            T->left= Insert(T->left,V);    }    return T;}Tree marktree(int n){    Tree t=(Tree)malloc(sizeof(struct TreeNode));    int V;    cin>>V;    t->v=V;    t->left=t->right=NULL;    t->flag=0;    for(int i=1;i<n;i++)    {        scanf("%d",&V);        t=Insert(t,V);    }    return t;}int check(Tree T,int V){    if(T->flag)    {        if(V > T->v )            return check(T->right,V);        else if(V < T->v )            return check(T->left,V);        else            return 0;    }    else    {        if(T->v == V)        {            T->flag=1;            return 1;        }        else        {            return 0;        }    }}int judge(Tree T,int N){    int V,flag=0;    cin>>V;    if(V!=T->v)        flag=1;    else T->flag=1;    for(int i=1;i<N;i++)    {        cin>>V;        if(!check(T,V)) flag=1;    }    if(flag)  return 0;    else      return 1;}void reset(Tree T){    if(T->right) reset(T->right);    if(T->left) reset(T->left);    T->flag=0;}void freetree(Tree T){    if(T->right) freetree(T->right);    if(T->left) freetree(T->left);    free(T);}int main(){    int N,L;    Tree T;    cin>>N;    while(N)    {        cin>>L;        T=marktree(N);        for(int i=0;i<L;i++)        {            if(judge(T,N)) cout<<"Yes"<<endl;            else cout<<"No"<<endl;            reset(T);        }        freetree(T);        cin>>N;    }}