【剑指offer6&18】：重建二叉树&:树的子结构

剑指offer6：重建二叉树

描述：输入某二叉树的前序遍历和中序遍历的结果，请重建出该二叉树。

例如二叉树的结构如下图所示

                                      

 

#include "stdafx.h"
#include<iostream>
#include<algorithm>
#include<string>
#include<vector>

using namespace std;

struct BinaryTreeNode
{
	int m_nValue;
	BinaryTreeNode* m_pLeft;
	BinaryTreeNode* m_pRight;

	BinaryTreeNode(int x) :m_nValue(x), m_pLeft(NULL), m_pRight(NULL) {}
};


BinaryTreeNode * ConstructCore(vector<int> pre, vector<int> vin)
{
	if (pre.empty() || vin.empty() || pre.size() != vin.size())
		return NULL;
	int index = 0;//保存中序遍历中根结点的位置参数
	vector<int> left_pre, left_vin, right_pre, right_vin;
	BinaryTreeNode* root = new BinaryTreeNode(pre[0]);

	//如果只有一个结点则直接返回
	if (pre.size() == 1)
	{
		if (vin.size() == 1 && pre[0] == vin[0])
			return root;
		else
		{
			cout << "Invalid input." << endl;
			return nullptr;
		}
			
	}


	//在中序遍历序列中找到根结点
	for (int i = 0; i < vin.size(); i++)
	{
		if (vin[i] == pre[0])
		{
			index = i;
			break;
		}
	}


	//分左右子树,见图分析
	for (int i = 0; i<index; i++)
	{
		//cout <<" m "<< endl;
		left_pre.push_back(pre[i + 1]);
		left_vin.push_back(vin[i]);
	}
	for (int i = index + 1; i<vin.size(); i++)
	{
		right_pre.push_back(pre[i]);
		right_vin.push_back(vin[i]);
	}
	//递归
	root->m_pLeft = ConstructCore(left_pre, left_vin);
	root->m_pRight = ConstructCore(right_pre, right_vin);

	return root;
}

//二叉树前序遍历打印
void PreOrderPrint(BinaryTreeNode* pHead)
{
	if (pHead == NULL)
		return;
	cout << pHead->m_nValue << "->";
	PreOrderPrint(pHead->m_pLeft);
	PreOrderPrint(pHead->m_pRight);
}
//二叉树的中序遍历打印
void InOrderPrint(BinaryTreeNode* pHead)
{
	if (pHead == NULL)
		return;
	InOrderPrint(pHead->m_pLeft);
	cout << pHead->m_nValue << "->";
	InOrderPrint(pHead->m_pRight);
}
//二叉树的后序遍历打印
void PostOrderPrint(BinaryTreeNode* pHead)
{
	if (pHead == NULL)
		return;
	PostOrderPrint(pHead->m_pLeft);
	PostOrderPrint(pHead->m_pRight);
	cout << pHead->m_nValue << "->";
}

int main()
{
	vector<int> Vinn{ 4,7,2,1,5,3,8,6 };
	vector<int> Pree{ 1,2,4,7,3,5,6,8 };
	//vector<int> Vinn{ 1 };//测试用例
	//vector<int> Pree{2};//测试用例
	BinaryTreeNode* newRoot = ConstructCore(Pree, Vinn);
	cout << "先序遍历：" << endl;
	PreOrderPrint(newRoot);
	cout<<endl << "中序遍历：" << endl;
	InOrderPrint(newRoot);
	cout<<endl << "后序遍历：" << endl;
	PostOrderPrint(newRoot);
	cout << endl;

	return 0;
}

测试结果：

 

剑指offer18：树的子结构

题目描述：输入两颗二叉树A和B，判断B是不是A的子结构。

解题技巧：

第一步，在树A中找到和B根结点值一样的结点R;

第二步，判断树A中以R为根结点的子树是不是包含和树B一样的结构

bool HasSubTree(BinaryTreeNode* pRoot1, BinaryTreeNode* pRoot2)
{
	if (pRoot1 == pRoot2)
		return true;
	if (pRoot2 == NULL)
		return true;
	if (pRoot1 == NULL)
		return false;
	if (pRoot1->m_nValue = pRoot2->m_nValue)
	{
		bool Lflag = HasSubTree(pRoot1->m_pLeft, pRoot2->m_pLeft);
		bool Rflag = HasSubTree(pRoot1->m_pRight, pRoot2->m_pRight);
		if (Lflag&&Rflag)
			return true;
	}
	bool Lflag = HasSubTree(pRoot1->m_pLeft, pRoot2);
	bool Rflag = HasSubTree(pRoot1->m_pRight, pRoot2);
	if (Lflag || Rflag)
	{
		return true;
	}
	else
		return false;
}