//面试题26：树的子结构

bool HasSubTree(BinaryTreeNode* pRoot1,BinaryTreeNode* pRoot2)

{

bool result=false;

if(pRoot1!=nullptr&&pRoot2!=nullptr)

{

if(Equal(pRoot1->m_dbValue,pRoot->m_dbValue))

result=DosTree1HaveTree2(pRoot1,pRoot2);

if(!result)

result=HasSubtree(pRoot1->m_pLeft,pRoot2);

if(!result)

result=HasSubtree(pRoot1->m_pRight,pRoot2);

}

return result;

}

bool DoesTree1HaveTree(BinaryTreeNode* pRoot1,BinaryTreeNode* pRoot2)

{

if(pRoot2==nullptr) return false;

if(pRoot1==nullptr) return false;

if(!Equal(pRoot1->m_dbValue,pRoot2->m_dbValue))

return false;

return DoesTree1HaveTree(pRoot1->m_pLeft,pRoot2->m_pLeft)&&DoesTree1HaveTree(pRoot1->m_pRight,pRoot2->m_pRight);

}

bool Equal(double num1,double num2)

{

if((num1-num2>-0.0000001) && (num1-num2<0.0000001))

return true;

else

return false;

}