将二叉查找树转化为链表的代码实现

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。

根据提供的思路，我自己写了一个读入任意序列的整数，建立二叉查找树再改成链表的C++代码:

#include <iostream>
#include <fstream>
#include <string>

using namespace std;

// binary search tree node
struct BSTreeNode
{
int value;
struct BSTreeNode *leftChild;
struct BSTreeNode *rightChild;
};

void addBSTreeNode(struct BSTreeNode **ppRoot/*double pointer to the root*/, int new_value);
void printBSTree(ofstream *fout, struct BSTreeNode *root, int level);
void treeToLinkedList(struct BSTreeNode **ppHead, struct BSTreeNode **ppTail, struct BSTreeNode *root);

int main() {
ifstream fin ("tree_to_linked_list.in");
ofstream fout ("tree_to_linked_list.out");
int in;
struct BSTreeNode *root = NULL, *head = NULL, *tail = NULL, *node = NULL;

if(fin.good() == false)
{
cerr << "file open failed" << endl;
return 0;
}

while(true)
{
if(fin.eof())
break;

fin >> in;

addBSTreeNode(&root, in);
}

printBSTree(&fout, root, 0);

fout << endl;

treeToLinkedList(&head, &tail, root);

for(node = head; node != NULL; node = node->rightChild)
{
fout << node->value << " ";
}
fout << endl;

for(node = tail; node != NULL; node = node->leftChild)
{
fout << node->value << " ";
}
fout << endl;

return 0;
}

void addBSTreeNode(struct BSTreeNode **ppRoot/*double pointer to the root*/, int new_value)
{
if(*ppRoot == NULL)
{
*ppRoot = new struct BSTreeNode();
(*ppRoot)->value = new_value;

return;
}

if(new_value == (*ppRoot)->value)
{
return;
}
else if(new_value < (*ppRoot)->value)
{
addBSTreeNode(&(*ppRoot)->leftChild, new_value);
}
else
{
addBSTreeNode(&(*ppRoot)->rightChild, new_value);
}
}

void printBSTree(ofstream *pFout, struct BSTreeNode *root, int level)
{
if(root == NULL)
return;

printBSTree(pFout, root->leftChild, level + 1);

int i;

for(i = 0; i < level; i ++)
(*pFout) << "\t";

(*pFout) << root->value << endl;

printBSTree(pFout, root->rightChild, level + 1);
}

void treeToLinkedList(struct BSTreeNode **ppHead, struct BSTreeNode **ppTail, struct BSTreeNode *root)
{
struct BSTreeNode *ltemp, *rtemp;

if(root == NULL)
{
(*ppHead) = NULL;
(*ppTail) = NULL;
return;
}

treeToLinkedList(ppHead, <emp, root->leftChild);
treeToLinkedList(&rtemp, ppTail, root->rightChild);

if(*ppHead == NULL)
{
(*ppHead) = root;
root->leftChild = NULL;
}
else
{
root->leftChild = ltemp;
ltemp->rightChild = root;
}

if(*ppTail == NULL)
{
(*ppTail) = root;
root->rightChild = NULL;
}
else
{
root->rightChild = rtemp;
rtemp->leftChild = root;
}
}

样例输入:

10 5 14 45 64 3 7 12 11 33 52 63 23 46 73 44 66 22 45 67 78 83

样例输出：

3
5
7
10
11
12
14
22
23
33
44
45
46
52
63
64
66
67
73
78
83

3 5 7 10 11 12 14 22 23 33 44 45 46 52 63 64 66 67 73 78 83
83 78 73 67 66 64 63 52 46 45 44 33 23 22 14 12 11 10 7 5 3