• 二元查找树转变成排序的双向链表


    转自http://zhedahht.blog.163.com/blog/static/254111742007127104759245/

    题目:输入一棵二元查找树,将该二元查找树转换成一个排序的双向链表。要求不能创建任何新的结点,只调整指针的指向。

      比如将二元查找树
                                                10
                                              /   
                                            6       14
                                          /       / 
                                       4     8  12    16
    转换成双向链表

    4=6=8=10=12=14=16。

    分析:本题是微软的面试题。很多与树相关的题目都是用递归的思路来解决,本题也不例外。下面我们用两种不同的递归思路来分析。

    思路一:当我们到达某一结点准备调整以该结点为根结点的子树时,先调整其左子树将左子树转换成一个排好序的左子链表,再调整其右子树转换右子链表。最近链接左子链表的最右结点(左子树的最大结点)、当前结点和右子链表的最左结点(右子树的最小结点)。从树的根结点开始递归调整所有结点。

    参考代码:
    
    首先我们定义二元查找树结点的数据结构如下:
        struct BSTreeNode // a node in the binary search tree
        {
            int          m_nValue; // value of node
            BSTreeNode  *m_pLeft;  // left child of node
            BSTreeNode  *m_pRight; // right child of node
        };
    
    思路一对应的代码:
    ///////////////////////////////////////////////////////////////////////
    // Covert a sub binary-search-tree into a sorted double-linked list
    // Input: pNode - the head of the sub tree
    //        asRight - whether pNode is the right child of its parent
    // Output: if asRight is true, return the least node in the sub-tree
    //         else return the greatest node in the sub-tree
    ///////////////////////////////////////////////////////////////////////
    BSTreeNode* ConvertNode(BSTreeNode* pNode, bool asRight)
    {
          if(!pNode)
                return NULL;
    
          BSTreeNode *pLeft = NULL;
          BSTreeNode *pRight = NULL;
    
          // Convert the left sub-tree
          if(pNode->m_pLeft)
                pLeft = ConvertNode(pNode->m_pLeft, false);
    
          // Connect the greatest node in the left sub-tree to the current node
          if(pLeft)
          {
                pLeft->m_pRight = pNode;
                pNode->m_pLeft = pLeft;
          }
    
          // Convert the right sub-tree
          if(pNode->m_pRight)
                pRight = ConvertNode(pNode->m_pRight, true);
    
          // Connect the least node in the right sub-tree to the current node
          if(pRight)
          {
                pNode->m_pRight = pRight;
                pRight->m_pLeft = pNode;
          }
    
          BSTreeNode *pTemp = pNode;
    
          // If the current node is the right child of its parent, 
          // return the least node in the tree whose root is the current node
          if(asRight)
          {
                while(pTemp->m_pLeft)
                      pTemp = pTemp->m_pLeft;
          }
          // If the current node is the left child of its parent, 
          // return the greatest node in the tree whose root is the current node
          else
          {
                while(pTemp->m_pRight)
                      pTemp = pTemp->m_pRight;
          }
     
          return pTemp;
    }
    
    ///////////////////////////////////////////////////////////////////////
    // Covert a binary search tree into a sorted double-linked list
    // Input: the head of tree
    // Output: the head of sorted double-linked list
    ///////////////////////////////////////////////////////////////////////
    BSTreeNode* Convert(BSTreeNode* pHeadOfTree)
    {
          // As we want to return the head of the sorted double-linked list,
          // we set the second parameter to be true
          return ConvertNode(pHeadOfTree, true);
    }
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  • 原文地址:https://www.cnblogs.com/theCambrian/p/3934574.html
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