判断树 B 是不是 树 A 的子树

#include <iostream>using namespace std;/** 描述：判断树 B 是否属于树 A 相当于，对树进行遍历的比较，当比较到树 B 的叶子为空时结束 注意：判断 树 A 和 树 B 是否为空的情况 注意这里生成树的规则是按照二叉树的前序遍历生成的，所以比较也要按照这个规则 **/typedef struct BiNode{    int value;    struct BiNode * rightChildNode;    struct BiNode * leftChildNode;}BiNode;typedef BiNode * BiLink;int Nil = 0;int nodeValue1[100]={1,2,4,0,6,0,0,5,0,7,0,0,3};int nodePoint1 = 0;int nodeValue2[100]={1,0,3};int nodePoint2 = 0;void initBiTree(BiLink * head,int type){    int value;    if(type == 1)        value = nodeValue1[nodePoint1++];    else        value = nodeValue2[nodePoint2++];    if(value == 0)        *head = NULL;    else{        *head = (BiLink)malloc(sizeof(BiNode));        (*head)->value = value;        if(type == 1){            initBiTree(&(*head)->leftChildNode,1);            initBiTree(&(*head)->rightChildNode,1);        }else{            initBiTree(&(*head)->leftChildNode,2);            initBiTree(&(*head)->rightChildNode,2);        }    }}// 遍历比较 树 A 和 树 B 的结点bool DoesTree1HaveTree2(BiLink T1 , BiLink T2){    // T2 为空证明遍历完毕    if(T2 == NULL)        return true;    if(T1 == NULL)        return false;    if(T1->value != T2->value)       return false;    return DoesTree1HaveTree2(T1->leftChildNode, T2->leftChildNode)&&        DoesTree1HaveTree2(T1->rightChildNode, T2->rightChildNode);}// 判断树 A 和 树 B 关系的函数bool HasSubTree(BiLink T1 , BiLink T2){    bool result = false;    if(T1 != NULL && T2 !=NULL){        if(T1->value == T2->value)            result = DoesTree1HaveTree2(T1,T2);        if(!result)            result = HasSubTree(T1->leftChildNode, T2);        if(!result)            result = HasSubTree(T1->rightChildNode, T2);    }    return result;}void PreOrderTraverse(BiLink T){    if(T == NULL)        return;    cout<<T->value<<" "<<endl;    PreOrderTraverse(T->leftChildNode);    PreOrderTraverse(T->rightChildNode);}int main(int argc, const char * argv[]) {    BiLink T1 = NULL;    BiLink T2 = NULL;    initBiTree(&T1, 1);    initBiTree(&T2, 2);    cout<<HasSubTree(T1, T2)<<endl;    return 0;}