把二元查找树转变成排序的双向链表

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。

　　思路二：我们可以中序遍历整棵树。按照这个方式遍历树，比较小的结点先访问。如果我们每访问一个结点，假设之前访问过的结点已经调整成一个排序双向链表，我们再把调整当前结点的指针将其链接到链表的末尾。当所有结点都访问过之后，整棵树也就转换成一个排序双向链表了。

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);
 }

思路二对应的代码：

///////////////////////////////////////////////////////////////////////
 // Covert a sub binary-search-tree into a sorted double-linked list
 // Input: pNode -           the head of the sub tree
 //        pLastNodeInList - the tail of the double-linked list
 ///////////////////////////////////////////////////////////////////////
 void ConvertNode(BSTreeNode* pNode, BSTreeNode*& pLastNodeInList)
 {
       if(pNode == NULL)
             return;
 
       BSTreeNode *pCurrent = pNode;
 
       // Convert the left sub-tree
       if (pCurrent->m_pLeft != NULL)
             ConvertNode(pCurrent->m_pLeft, pLastNodeInList);
 
       // Put the current node into the double-linked list
       pCurrent->m_pLeft = pLastNodeInList; 
       if(pLastNodeInList != NULL)
             pLastNodeInList->m_pRight = pCurrent;
 
       pLastNodeInList = pCurrent;
 
       // Convert the right sub-tree
       if (pCurrent->m_pRight != NULL)
             ConvertNode(pCurrent->m_pRight, pLastNodeInList);
 }
 
 ///////////////////////////////////////////////////////////////////////
 // Covert a binary search tree into a sorted double-linked list
 // Input: pHeadOfTree - the head of tree
 // Output: the head of sorted double-linked list
 ///////////////////////////////////////////////////////////////////////
 BSTreeNode* Convert_Solution1(BSTreeNode* pHeadOfTree)
 {
       BSTreeNode *pLastNodeInList = NULL;
       ConvertNode(pHeadOfTree, pLastNodeInList);
 
       // Get the head of the double-linked list
       BSTreeNode *pHeadOfList = pLastNodeInList;
       while(pHeadOfList && pHeadOfList->m_pLeft)
             pHeadOfList = pHeadOfList->m_pLeft;
 
       return pHeadOfList;
 }

以上转自何海涛博客

下面贴一个LEETCODE上的代码，最后的返回结果是一个环形链表，有一个头指针指向最小元素。

// This is a modified in-order traversal adapted to this problem.
// prev (init to NULL) is used to keep track of previously traversed node.
// head pointer is updated with the list's head as recursion ends.
void treeToDoublyList(Node *p, Node *& prev, Node *& head) {
	if (!p) return;
	treeToDoublyList(p->left, prev, head);
	// current node's left points to previous node
	p->left = prev;
	if (prev)
		prev->right = p;  // previous node's right points to current node
	else
		head = p; // current node (smallest element) is head of
	// the list if previous node is not available
	// as soon as the recursion ends, the head's left pointer
	// points to the last node, and the last node's right pointer
	// points to the head pointer.
	Node *right = p->right;
	head->left = p;
	p->right = head;
	// updates previous node
	prev = p;
	treeToDoublyList(right, prev, head);
}

// Given an ordered binary tree, returns a sorted circular
// doubly-linked list. The conversion is done in-place.
Node* treeToDoublyList(Node *root) {
	Node *prev = NULL;
	Node *head = NULL;
	treeToDoublyList(root, prev, head);
	return head;
}