练习系列 - 1、把二元查找树转变成排序的双向链表

/*!
\author LiuBao
\date 2011/2/25
\brief 把二元查找树转变成排序的双向链表
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。
*/
#include <stdio.h>
#include <stdlib.h>
 
struct BSTreeNode                   ///  二元查找树/双向链表节点
{
    int m_nValue;                   ///< 节点值
    struct BSTreeNode *m_pLeft;     ///< 左儿子指针/前节点指针
    struct BSTreeNode *m_pRight;    ///< 右儿子指针/后节点指针
};
 
enum ChildPosition                  ///  儿子节点插入位置
{
    Left,                           ///< 插入为左儿子
    Right                           ///< 插入为右儿子
};
 
/*!
创建值为child_value的节点,并插入为parent的pos儿子
\param parent 待插入儿子节点的父节点指针
\param pos 儿子节点插入位置
\param child_value 儿子节点的值
\return 返回新插入的儿子节点的指针
*/
struct BSTreeNode *add_child_to_tree(struct BSTreeNode *parent,
                                     enum ChildPosition pos,
                                     int child_value)
{
    struct BSTreeNode *retVal = NULL;
 
    if(parent)
    {
        struct BSTreeNode *child = malloc(sizeof(struct BSTreeNode));
 
        if(child)
        {
            child->m_nValue = child_value;
            child->m_pLeft = NULL;
            child->m_pRight = NULL;
 
            if(pos == Left)
                parent->m_pLeft = child;
            else
                parent->m_pRight = child;
 
            retVal = child;
        }
    }
 
    return retVal;
}
 
/*!
创建一个二元查找树,并返回树根指针
\return 二元查找树的根指针
*/
struct BSTreeNode *create_tree()
{
    struct BSTreeNode *root = malloc(sizeof(struct BSTreeNode));
 
    if(root)
    {
        root->m_nValue = 10;
        root->m_pLeft = NULL;
        root->m_pRight = NULL;
    }
 
    {
        struct BSTreeNode *lChild = add_child_to_tree(root, Left, 6);
        struct BSTreeNode *rChild = add_child_to_tree(root, Right, 14);
 
        (void)add_child_to_tree(lChild, Left, 4);
        (void)add_child_to_tree(lChild, Right, 8);
        (void)add_child_to_tree(rChild, Left, 12);
        (void)add_child_to_tree(rChild, Right, 16);
    }
 
    return root;
}
 
/*!
从二元查找树根开始,递归中根遍历转换其为有序双向链表(不可重入)
\param root 二元查找树根
*/
void change_ptr(struct BSTreeNode *root)
{
    static struct BSTreeNode *last = NULL;
 
    if(root)
    {
        change_ptr(root->m_pLeft);
        if(last)
        {
            last->m_pRight = root;
            root->m_pLeft = last;
        }
        last = root;
        change_ptr(root->m_pRight);
    }
}
 
/*!
从curr节点找到头结点,并返回头结点指针
\param curr 当前节点
\return 双向链表头结点指针
*/
struct BSTreeNode *find_head(struct BSTreeNode *curr)
{
    struct BSTreeNode *head = NULL;
 
    while(curr)
    {
        head = curr;
        curr = curr->m_pLeft;
    }
 
    return head;
}
 
/*!
把根节点为root的二元查找树转换为有序双向链表,并返回链表头指针
\param root 二元查找树的根节点
\return 转换为的双向有序链表头指针
*/
struct BSTreeNode *to_link_list(struct BSTreeNode *root)
{
    change_ptr(root);
    return find_head(root);
}
 
int main()
{
    struct BSTreeNode *bTree = create_tree();
    struct BSTreeNode *head = to_link_list(bTree);
 
    /* 测试双向有序链表 */
    {
        struct BSTreeNode *tail = NULL;
 
        while(head)
        {
            printf("%d\n", head->m_nValue);
            tail = head;
            head = head->m_pRight;
        }
 
        printf("reverse iterate:\n");
 
        while(tail)
        {
            printf("%d\n", tail->m_nValue);
            tail = tail->m_pLeft;
        }
    }
 
    return 0;
}
posted @ 2011-02-25 11:34  紫红的泪  阅读(2146)  评论(0编辑  收藏  举报