树的子结构

来源：互联网发布：js字符串是一个数组编辑：程序博客网时间：2024/05/17 02:15

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：

先在Ａ中找Ｂ的跟节点，若找到相同的，在判断左右子树是否相同．

struct BinaryTreeNode{    int     m_nValue;    BinaryTreeNode* m_pLeft;    BinaryTreeNode* m_pRight;    BinaryTreeNode(int n) : m_nValue(n), m_pLeft(NULL), m_pRight(NULL) {};};

bool HasSubtree(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)        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);}

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