// test20.cpp : 定义控制台应用程序的入口点。
//
#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.先在二叉树tree1中找到tree2的根节点
//2.在判断tree2的左右孩子是否在tree1中
//3.如果不存在继续遍历tree1的左孩子和右孩子
bool HasSubtree(TreeNode* pRoot1, TreeNode* pRoot2)
{
if (pRoot1 == NULL) return false;//这一点很重要，要不然if (pRoot1->val == pRoot2->val)会报空指针的错误
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为空，返回false;如果tree2为空，返回true；如果两者都不为空，继续判断其左右孩子
bool DoseTree1HasTree2(TreeNode* tree1, TreeNode* tree2)
{
if (tree1 == NULL &&tree2 == NULL) return true;//tree1和tree2都为空
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 << "创建T1:" << endl;
so.preCreate(T1);
cout << "创建T1成功！" << endl;
cout << "创建T2:" << endl;
so.preCreate(T2);
cout << "创建T2成功！" << endl;
cout << "T1的前序遍历:" << endl;
so.preOrder(T1);
cout << endl;
//cout << "T2的前序遍历:" << endl;
//so.preOrder(T2);
//cout << endl;
cout << "T2是不是T1的子树：" << endl;
bool result = so.HasSubtree(T1,T2);
cout << result << endl;
cout << endl;
return 0;
}