pta-子孙关系判断 (50 分)

已知二叉树的先序序列，判断结点u是否是结点v的子孙，是就输出v到u的路径长度，否则输出NO。假设结点个数少于50个。

输入格式:

输入共二行，第一行中给出先序序列，第二行给出两个顶点。*表示空树。

输出格式:

输出一个整数或NO。

输入样例1:

ABC**DE*G**F***
BE

结尾无空行

输出样例1:

2

结尾无空行

本人用的是C++模版写的，但是这并不能够影响本题的本质，所以，仅看代码内容即可。

//创建一棵二叉树（先序顺序）

//构造函数
template 
BiNode *BiTree::Creat(BiNode *bt){
    ElemType ch;
//    cout << "请输入创建一棵二叉树的节点数据：" << endl;
    cin >> ch;
    if (ch == '*'){
        return NULL;
    }
    else{
        bt = new BiNode;
        bt->data = ch;
        bt->lchild = Creat(bt->lchild);
        bt->rchild = Creat(bt->rchild);
    }
    return bt;
}

//找到首节点，并且计算路径

//计算路径
int length = -1;        //这里是全局变量
template 
int BiTree::CountPath(BiNode *root, ElemType data2){
    if (root == NULL) {
        length++;
        return 0;
    }
    else{
        length++;
        if (root->data == data2) {
            return length;
        }
        if(!CountPath(root->lchild, data2)) length--;
        else return length;
        if(!CountPath(root->rchild, data2)) length--;
        else return length;
        return 0;
    }
}
//找到首节点
template 
int BiTree::FindTree(BiNode *bt, ElemType data1, ElemType data2){
    if (bt == NULL) {
        return 0;
    }
    else{
        if (bt->data == data1) {
            return CountPath(bt, data2);
        }
        if(FindTree(bt->lchild, data1, data2)) return length;
        if(FindTree(bt->rchild, data1, data2)) return length;
    }
    return 0;
}

下面附上所有的代码

#include 
#include 
#include 
using namespace std;

template 
struct BiNode{
    ElemType data;
    BiNode  *lchild,*rchild;
};

int length = -1;

template
class BiTree{
private:
    BiNode *root;
    BiNode *Creat(BiNode *bt);
    void Release(BiNode *bt);
    void InOrder(BiNode *bt);
    int Depth(BiNode *bt);
    int FindTree(BiNode *bt,ElemType data1,ElemType data2);
    int CountPath(BiNode *root,ElemType data2);
public:
    BiTree(){root = Creat(root);}
    ~BiTree(){
        Release(root);
    }
    void InOrder(){
        InOrder(root);
    }
    int Depth(){
        return Depth(root);
    }
    
    int CountPath(ElemType data1,ElemType data2){
        return FindTree(root, data1, data2);
    }
};
//寻找并计算路径
template 
int BiTree::CountPath(BiNode *root, ElemType data2){
    if (root == NULL) {
        length++;
        return 0;
    }
    else{
        length++;
        if (root->data == data2) {
            return length;
        }
        if(!CountPath(root->lchild, data2)) length--;
        else return length;
        if(!CountPath(root->rchild, data2)) length--;
        else return length;
        return 0;
    }
}
//找到首节点
template 
int BiTree::FindTree(BiNode *bt, ElemType data1, ElemType data2){
    if (bt == NULL) {
        return 0;
    }
    else{
        if (bt->data == data1) {
            return CountPath(bt, data2);
        }
        if(FindTree(bt->lchild, data1, data2)) return length;
        if(FindTree(bt->rchild, data1, data2)) return length;
    }
    return 0;
}

//构造函数
template 
BiNode *BiTree::Creat(BiNode *bt){
    ElemType ch;
//    cout << "请输入创建一棵二叉树的节点数据：" << endl;
    cin >> ch;
    if (ch == '*'){
        return NULL;
    }
    else{
        bt = new BiNode;
        bt->data = ch;
        bt->lchild = Creat(bt->lchild);
        bt->rchild = Creat(bt->rchild);
    }
    return bt;
}

//析构函数
template 
void BiTree::Release(BiNode *bt){
    if (bt != NULL) {
        Release(bt->lchild);
        Release(bt->rchild);
        delete bt;
    }
}

//求二叉树的深度
template 
int BiTree::Depth(BiNode *bt){
    if (bt == NULL) {
        return 0;
    }
    else{
        int dep1 = Depth(bt->lchild);
        int dep2 = Depth(bt->rchild);
        return (dep1>dep2)?(dep1+1):(dep2+1);
    }
}

//中序遍历
template 
void BiTree::InOrder(BiNode *bt){
    if(bt==NULL){
        return;
    }
    else{
        InOrder(bt->lchild);
        cout << bt->data << " ";
        InOrder(bt->rchild);
    }
}

int main(int argc, const char * argv[]) {
    char tr1,tr2;
    BiTree T;
    cin >> tr1 >>tr2 ;
    if(T.CountPath(tr1, tr2)){
        cout << length ;
    }
    else cout << "NO";
    return 0;
}