面试题16:树的子结构

2015-11-21 01:27:45 · 作者: · 浏览: 8
#include "stdafx.h"   
#include    
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 nNodeva lue = 0;  
    cin >> nNodeva lue;      
    if (-1 == nNodeva lue)  
    {  
       return;   
    }  
    else  
    {  
        pRoot = new BinaryTreeNode();  
        pRoot->m_nValue = nNodeva lue;  
        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;  
}  
#include "stdafx.h"
#include 
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 nNodeva lue = 0; cin >> nNodeva lue; if (-1 == nNodeva lue) { return; } else { pRoot = new BinaryTreeNode(); pRoot->m_nValue = nNodeva lue; 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; }

\