剑指offer-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。

思路：第一步在树中找到和B的根结点的值一样的结点R，第二步再判断树A中以R为根结点的子树是不是包含B一样的结构。

bool HasSubTree(BinaryTreeNode* pRoot1, BinaryTreeNode* pRoot2)
{
    bool result = false;
    if(pRoot1 != NULL && pRoot1 != NULL)
    {
        if(pRoot1->m_nValue == pRoot->m_nValue)
            result = DoesTree1HaveTree2(pRoot1, pRoot2);
        if(!result)
            result = HasSubTree(pRoot->m_pLeft, pRoot2);
        if(!result)
            result = HasSubTree(pRoot->m_pRight, pRoot2);
    }

    return result;
}

bool DoesTree1HaveTree2(BinaryTreeNode* pRoot1, BinaryTreeNode* pRoot2)
{
    //注意pRoot2要在pRoot1前判空
    if(pRoot2 == NULL)
        return true;
    
    if(pRoot1 == NULL)
        return false;
    
    if(pRoot1->m_nValue != pRoot->m_nValue)
        return false;

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