剑指offer刷题之c++实现的树的子结构

#include "myHead.h"
 #include "allConstructBinaryTree.cpp" 
#include <string.h>
/*
输入两颗二叉树A，B，判断B是不是A的子结构。
*/
  void inOrder(TreeNode *h,vector<int> &res);
  TreeNode* createBinaryTree1();
  TreeNode* createBinaryTree2();
  bool HasSubtree(TreeNode* pRoot1, TreeNode* pRoot2)
{
    if(pRoot2 == NULL)
	return false;
	if(pRoot1 ==NULL)
	return false;
	vector<int> res1,res2;
	string s1="",s2="";
	const	char *c1,*c2;
	char t ='0'; 
	inOrder(pRoot1,res1);
	for(vector<int>::iterator it=res1.begin();it<res1.end();++it){
		  t=(*it)+t;
		  s1=s1+t;
		  t='0';
	}
	inOrder(pRoot2,res2);
	 t='0';
	for(vector<int>::iterator it=res2.begin();it<res2.end();++it){
		  t=(*it)+t;
		  s2=s2+t;
		  t='0';
	}
//将string类型的转化为char*来实现自定义的操作，C++标准库也为了和之前用C写的程序兼容，于是可以用string的c_str()函数。
//string var = "Olympic";
//char *ptr = var.c_str(); //还不能被编译
//但是c_str()为了防止意外地修改string对象，返回的是const指针，所以上面这段代码是不能被编译的。正确的应该是用const指针。
//string var = "Olympic";
//const char *p = var.c_str(); //Correct!
	c1 = s1.c_str();
	c2 = s2.c_str();
	
	return strstr(c1,c2);
}
 void inOrder(TreeNode *h,vector<int> &res){
 	 if(h){
 	 	    res.push_back(h->val);
 		    inOrder(h->left,res);
//	 		cout<<h->val;	
	 		inOrder(h->right,res);
	  }
 }
 TreeNode* createBinaryTree1(){
//    	TreeNode *T = (TreeNode *) malloc(sizeof(TreeNode)) ;
//		T->val = 8;
//		  T->left = (TreeNode *) malloc(sizeof(TreeNode)) ;
//		  T->left->val = 8;
//		    T->left->left = (TreeNode *) malloc(sizeof(TreeNode)) ;
//			T->left->left->val = 9;
//			T->left->left->left = T->left->left->right = NULL;
//			T->left->right = (TreeNode *) malloc(sizeof(TreeNode)) ;
//			T->left->right->val = 2;
//			T->left->right->left = (TreeNode *) malloc(sizeof(TreeNode)) ;
//			T->left->right->left->val = 4;
//			T->left->right->left->left=T->left->right->left->right =NULL;
//			T->left->right->right = (TreeNode *) malloc(sizeof(TreeNode)) ;
//			T->left->right->right->val = 7;
//			T->left->right->right->left =T->left->right->right->right = NULL;
//		  T->right = (TreeNode *) malloc(sizeof(TreeNode)) ;
//		  T->right->val = 7;	
//		  	 T->right->left = T->right->right=NULL ;
			TreeNode *T = (TreeNode *) malloc(sizeof(TreeNode)) ;
		T->val = 8;
		  T->left = (TreeNode *) malloc(sizeof(TreeNode)) ;
		  T->left->val = 9;
		    T->left->left =  T->left->right = NULL;
		  T->right = (TreeNode *) malloc(sizeof(TreeNode)) ;
		  T->right->val = 3;	
		  	 T->right->left == (TreeNode *) malloc(sizeof(TreeNode)) ;
		  T->right->left->val = 2; 
			   T->right->right = NULL;	 
			return T;		
    } 
TreeNode* createBinaryTree2(){
    	TreeNode *T = (TreeNode *) malloc(sizeof(TreeNode)) ;
		T->val = 8;
		  T->left = (TreeNode *) malloc(sizeof(TreeNode)) ;
		  T->left->val = 9;
		    T->left->left =  T->left->right = NULL;
		  T->right = (TreeNode *) malloc(sizeof(TreeNode)) ;
		  T->right->val = 2;	
		  	 T->right->left = T->right->right = NULL;
			return T;		
    } 
 int main(){
	 TreeNode* h1 =	createBinaryTree1();
	 TreeNode* h2 =NULL;//=	createBinaryTree2();
	 HasSubtree(h1,h2);
 }