1.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。
    10
     / \
  6   14
 / \    / \
4 8 12 16
转换成双向链表
4=6=8=10=12=14=16。
首先我们定义的二元查找树节点的数据结构如下：

struct BSTreeNode
{
int m_nValue; // value of node
BSTreeNode *m_pLeft; // left child of node
BSTreeNode *m_pRight; // right child of node
};

HANDWRITING:

BSTreeNode *list;
BSTreeNode *tree;
BSTreeNode *transform (BSTreeNode * & head, BSTreeNode * & tail, BSTreeNode *root) {
if (root == 0) return 0;
BSTreeNode *ltail, *rhead;
transform(head, ltail, root->m_pLeft);
transform(rhead, tail, root->mpRight);
if (root == tree) list = head;
if (ltail == 0)  head = root;
else ltail.m_pRight = root, root.m_pLeft = ltail;
if (rhead == 0) tail = root;
else root.m_pRight = rhead, rhead.m_pLeft = root;
}

ANSWER FROM:http://blog.csdn.net/v_july_v/article/details/6870251

This is a traditional problem that can be solved using recursion. 

For each node, connect the double linked lists created from left and right child node to form a full list.

/**
* @param root The root node of the tree
* @return The head node of the converted list.
*/
BSTreeNode * treeToLinkedList(BSTreeNode * root) {
BSTreeNode * head, * tail;
helper(head, tail, root);
return head;
}
void helper(BSTreeNode *& head, BSTreeNode *& tail, BSTreeNode *root) {
BSTreeNode *lt, *rh;
if (root == NULL) {
head = NULL, tail = NULL;
return;
}
helper(head, lt, root->m_pLeft);
helper(rh, tail, root->m_pRight);
if (lt!=NULL) {
lt->m_pRight = root;
root->m_pLeft = lt;
} else  {
head = root;
}
if (rh!=NULL) {
root->m_pRight=rh;
rh->m_pLeft = root;
} else {
tail = root;
}
}