面试题16：树的子结构

最新推荐文章于 2023-02-12 10:58:56 发布

龙少爷_阿呆

最新推荐文章于 2023-02-12 10:58:56 发布

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分类专栏：面试题 C++

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本文介绍了一种算法，用于判断一棵二叉树是否包含另一棵二叉树的结构。通过递归比较两棵树的节点值，实现了子结构的查找，并提供了完整的C++代码实现。

思路：首先在树A中找到和B的根结点的值一样的结点R，然后再判断树A中以R为根的子树是否包含树B一样的结构。

递归代码：

#include "stdafx.h"
#include <iostream>
using namespace std;

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

bool DoesTree1haveTree2(BinaryTreeNode *pRoot1, BinaryTreeNode *pRoot2)
{
	if (pRoot2 == NULL)
	{
		return true;
	}

	if (pRoot1 == NULL)
	{
		return false;
	}

	if (pRoot1->m_nValue != pRoot2->m_nValue)
	{
		return false;
	}

	return DoesTree1haveTree2(pRoot1->m_pLeft, pRoot2->m_pLeft)
		&& DoesTree1haveTree2(pRoot1->m_pRight, pRoot2->m_pRight);
}

//判断B是否为A的子结构
bool IsSubStruct(BinaryTreeNode *pRootOfTreeA, BinaryTreeNode *pRootOfTreeB)
{
   bool result = false;
   if (pRootOfTreeA != NULL && pRootOfTreeB != NULL)
   {
	   if (pRootOfTreeA->m_nValue == pRootOfTreeB->m_nValue)
	   {
		   result = DoesTree1haveTree2(pRootOfTreeA, pRootOfTreeB);
	   }

	   if (!result)
	   {
		   result = IsSubStruct(pRootOfTreeA->m_pLeft, pRootOfTreeB);
	   }

	   if (!result)
	   {
		   result = IsSubStruct(pRootOfTreeA->m_pRight, pRootOfTreeB);
	   }
   } 

   return result;
}

//以先序的方式构建二叉树，输入-1表示结点为空
void CreateBinaryTree(BinaryTreeNode *&pRoot)
{
	int nNodeValue = 0;
	cin >> nNodeValue;	
	if (-1 == nNodeValue)
	{
       return; 
	}
	else
	{
		pRoot = new BinaryTreeNode();
		pRoot->m_nValue = nNodeValue;
		CreateBinaryTree(pRoot->m_pLeft);
		CreateBinaryTree(pRoot->m_pRight);
	}
}

void PrintInOrder(BinaryTreeNode *&pRoot)
{
	if (pRoot != NULL)
	{
		PrintInOrder(pRoot->m_pLeft);
		cout << pRoot->m_nValue << " ";
		PrintInOrder(pRoot->m_pRight);
	}
}

int _tmain(int argc, _TCHAR* argv[])
{
	BinaryTreeNode *pRoot1 = NULL;
	CreateBinaryTree(pRoot1);
	PrintInOrder(pRoot1);
	cout << endl;

	BinaryTreeNode *pRoot2 = NULL;
	CreateBinaryTree(pRoot2);
	PrintInOrder(pRoot2);
	cout << endl;

	cout << "Root2是否为Root1的子结构，是返回1，否则返回0.结果是： " << IsSubStruct(pRoot1, pRoot2) << endl;

	system("pause");
	return 0;
}