剑指Offer系列-面试题18：树的子结构

题目：输入两颗二叉树A和B，判断B是不是A的子结构。二叉树结点的定义如下：

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

思路：需要多次判断指针为NULL时的情况，这也是递归结束的标识。

代码：

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);
}

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;
}