面试题18：树的子结构

掌握这题的思路：

第一步在树A中找到和B的根节点的值一样的节点R，第二步在判断树A中以R为根节点的子树是不是包含和树B一样的结构

bool HasSubtreeCore(BinaryTreeNode* pRoot1, BinaryTreeNode* pRoot2);
bool DoesTree1HaveTree2(BinaryTreeNode* pRoot1, BinaryTreeNode* pRoot2);

bool HasSubtree(BinaryTreeNode* pRoot1, BinaryTreeNode* pRoot2)
{
    bool result = false;

    if(pRoot1 != NULL && pRoot2 != NULL)
    {
        if(pRoot1->m_nValue == pRoot2->m_nValue)
            result = DoesTree1HaveTree2(pRoot1, pRoot2);
        if(!result)
            result = HasSubtree(pRoot1->m_pLeft, pRoot2);
        if(!result)
            result = HasSubtree(pRoot1->m_pRight, pRoot2);
    }

    return result;
}

bool DoesTree1HaveTree2(BinaryTreeNode* pRoot1, BinaryTreeNode* pRoot2)
{
    if(pRoot2 == NULL)    //注意判断和树2相同的条件是把树2都遍历完
        return true;

    if(pRoot1 == NULL)
        return false;

    if(pRoot1->m_nValue != pRoot2->m_nValue)    //这里用!=要是用==就变得复杂了
        return false;

    return DoesTree1HaveTree2(pRoot1->m_pLeft, pRoot2->m_pLeft) &&
        DoesTree1HaveTree2(pRoot1->m_pRight, pRoot2->m_pRight);
}