求解(树的子结构问题-何海涛100题)

来源：互联网发布：菲律宾进出口数据编辑：程序博客网时间：2024/05/17 08:19

下面的题目是看何海涛100题中遇到的，自己看不懂，特贴出来请大家帮忙。原文地址为http://zhedahht.blog.163.com/blog/static/25411174201011445550396/

http://zhedahht.blog.163.com/

题目：二叉树的结点定义如下：

struct TreeNode

{

int m_nValue;

TreeNode* m_pLeft;

TreeNode* m_pRight;

};

输入两棵二叉树A和B，判断树B是不是A的子结构。

例如，下图中的两棵树A和B，由于A中有一部分子树的结构和B是一样的，因此B就是A的子结构。

                 1                                                   8
               /    \                                               /    \
              8    7                                             9    2
            /    \
           9    2
                / \
               4  7

bool HasSubtree(TreeNode* pTreeHead1, TreeNode* pTreeHead2)

{

if((pTreeHead1 == NULL && pTreeHead2 != NULL) ||

(pTreeHead1 != NULL && pTreeHead2 == NULL))

return false;

if(pTreeHead1 == NULL && pTreeHead2 == NULL)

return true;

return HasSubtreeCore(pTreeHead1, pTreeHead2);

}

bool HasSubtreeCore(TreeNode* pTreeHead1, TreeNode* pTreeHead2)

{

bool result = false;

if(pTreeHead1->m_nValue == pTreeHead2->m_nValue)

{

result = DoesTree1HaveAllNodesOfTree2(pTreeHead1, pTreeHead2);

}

if(!result && pTreeHead1->m_pLeft != NULL)

result = HasSubtreeCore(pTreeHead1->m_pLeft, pTreeHead2);

if(!result && pTreeHead1->m_pRight != NULL)

result = HasSubtreeCore(pTreeHead1->m_pRight, pTreeHead2);

return result;

}

bool DoesTree1HaveAllNodesOfTree2(TreeNode* pTreeHead1, TreeNode* pTreeHead2)

{

if(pTreeHead2 == NULL)【此处是什么意思？】

return true;

if(pTreeHead1 == NULL)【此处是什么意思？】

return false;

if(pTreeHead1->m_nValue != pTreeHead2->m_nValue)

return false;

return DoesTree1HaveAllNodesOfTree2(pTreeHead1->m_pLeft, pTreeHead2->m_pLeft) &&

DoesTree1HaveAllNodesOfTree2(pTreeHead1->m_pRight, pTreeHead2->m_pRight);

}

接触C++虽然有一段时间了，但是上面的程序依然有地方看不懂，请师兄师姐学者指导！

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