輸入兩棵二叉樹A，B，判斷B是不是A的子結構。（ps：我們約定空樹不是任意一個樹的子結構）

// test20.cpp : 定義控制台應用程序的入口點。
//

#include "stdafx.h"
#include<iostream>
#include<vector>
#include<string>
#include<queue>
#include<stack>

using namespace std;



struct TreeNode {
	int val;
	struct TreeNode *left;
	struct TreeNode *right;
	TreeNode(int x) :
	val(x), left(NULL), right(NULL) {
	}
};
class Solution {
public:
	//1.先在二叉樹tree1中找到tree2的根節點
	//2.在判斷tree2的左右孩子是否在tree1中
	//3.如果不存在繼續遍歷tree1的左孩子和右孩子
	bool HasSubtree(TreeNode* pRoot1, TreeNode* pRoot2)
	{
		if (pRoot1 == NULL) return false;//這一點很重要，要不然if (pRoot1->val == pRoot2->val)會報空指針的錯誤
		if (pRoot2 == NULL) return false;

		bool result = false;
		if (pRoot1->val == pRoot2->val)
		{
			result = DoseTree1HasTree2(pRoot1,pRoot2);
		}
		if (result == false)
			result = HasSubtree(pRoot1->left,pRoot2);
		if (result == false)
			result = HasSubtree(pRoot1->right,pRoot2);
		return result;

	}

	//如果tree1為空，返回false;如果tree2為空，返回true；如果兩者都不為空，繼續判斷其左右孩子
	bool DoseTree1HasTree2(TreeNode* tree1, TreeNode* tree2)
	{
		if (tree1 == NULL &&tree2 == NULL) return true;//tree1和tree2都為空
		else if (tree1 == NULL && tree2!=NULL) return false;
		else if (tree1 != NULL && tree2==NULL) return true;
		else 
		{
			if (tree1->val != tree2->val) return false;
			return DoseTree1HasTree2(tree1->left, tree2->left) && DoseTree1HasTree2(tree1->right, tree2->right);
		}
		
	}



	void preCreate(TreeNode* &T)
	{
		int num;
		cin >> num;
		if (num == 0) T = NULL;
		else
		{
			T = new TreeNode(num);
			preCreate(T->left);
			preCreate(T->right);
		}
	}

	void preOrder(TreeNode* T)
	{
		if (T == NULL) return; 
		else
		{
			cout << T->val << "  ";
			preOrder(T->left);
			preOrder(T->right);
		}
	}
};
int main()
{
	
	Solution so;
	TreeNode *T1;
	TreeNode *T2;
	vector<int> pre = { 1,2,4,7,3,5,6,8 };
	vector<int> in = { 4,7,2,1,5,3,8,6 };

	cout << "創建T1:" << endl;
	so.preCreate(T1);
	cout << "創建T1成功！" << endl;


	cout << "創建T2:" << endl;
	so.preCreate(T2);
	cout << "創建T2成功！" << endl;

	cout << "T1的前序遍歷:" << endl;
	so.preOrder(T1);
	cout << endl;

	//cout << "T2的前序遍歷:" << endl;
	//so.preOrder(T2);
	//cout << endl;

	cout << "T2是不是T1的子樹：" << endl;
	bool result = so.HasSubtree(T1,T2);

	cout << result << endl;



	
	
	cout << endl;
	return 0;
}