二叉查找树(续)

题记

之前对二叉查找树的基本操作写了一篇文章《二叉查找树》，觉得还不过瘾。毕竟还没有实践。所以找了几个笔试题做做。小结如下：

经典题1：

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，二叉查找树的结构为

                      10
                    /    \
                  6       14
                /  \     /　 \
            　4     8  12 　  16

调整后的结果是  4=6=8=10=12=14=16

经典题2：

输入一颗二元查找树，将该树转换为它的镜像，即在转换后的二元查找树中，左子树的结点都大于右子树的结点。用递归和循环两种方法完成树的镜像转换。

比如

                      10
                    /    \
                  6       14
                /  \     /　 \
            　4     8  12 　  16

变成

                      10
                    /    \
                  14      6
                /  \    /　 \
            　16   12  8 　  4

以上两题的二叉树结构都是

struct BSTreeNode // a node in the binary search tree (BST){      int          m_nValue; // value of node      BSTreeNode  *m_pLeft;  // left child of node      BSTreeNode  *m_pRight; // right child of node};

请先自己试着解答，然后看下面的分析和代码！【注意，本代码最后没有对bstree的内存释放！仅做参考】

题一：分析部分请见注释！！

#include<stdio.h>#include<stdlib.h>#include<assert.h>typedef struct BSTreeNode{int m_value;struct BSTreeNode *m_pLeft;struct BSTreeNode *m_pRight;}BSTreeNode;//生成二叉查找树BSTreeNode *insert(BSTreeNode *tree,BSTreeNode *n){if(tree == NULL){tree = n;}else if(n->m_value < tree->m_value){tree->m_pLeft = insert(tree->m_pLeft,n);}else{tree->m_pRight = insert(tree->m_pRight,n);}return tree;}//方法一:递归思想。//假设节点左右子树都已经是"已排序的双向链表"，比如左子树链表指针为tLeft,右子树链表指针为tRight//那么，只需要把此时的节点加入到这个链表里，最后返回指向这个调整后的链表即可！//参数含义：//cnode：当前处理的节点//asLeft: 值取-1,0或1。-1表示无父节点;1表示当前节点是其父节点的左孩子;0表示是其父节点的右孩子；//之所以asLeft这样取值，是因为最后返回的指针会指向链表的尾部或头部(尾部是cnode的前驱，头部是cnode的后继)！需要用它来判断！//连接左孩子时需要链表的尾指针(tLeft)，连接右孩子时需要链表的头指针(tRight)！BSTreeNode * convert(BSTreeNode *cnode,int asLeft){if(cnode == NULL){return NULL;}BSTreeNode *tLeft=NULL, *tRight=NULL;if(cnode->m_pLeft != NULL){tLeft = convert(cnode->m_pLeft,1);}if(tLeft != NULL){//调整好左子树cnode->m_pLeft = tLeft;tLeft->m_pRight = cnode;//左指针指向前驱节点，右指针指向后继节点}if(cnode->m_pRight != NULL){tRight = convert(cnode->m_pRight,0);}if(tRight != NULL){//调整好右子树cnode->m_pRight = tRight;tRight->m_pLeft = cnode;}//以上部分已经调整好左右子树了，下面来返回该链表指针if(asLeft == 0 || asLeft == -1){//如果没有父亲，则让其指向最小节点！while(cnode->m_pLeft != NULL){cnode = cnode->m_pLeft;}}else{//如果以cnode为根的子树是其父亲的左子树，那么父亲的前驱就是这颗子树中最大的节点while(cnode->m_pRight != NULL){cnode = cnode->m_pRight;}}return cnode;}//方法二：还是递归法//区别方法一：本方法利用中序遍历，每次递归时只假设当前节点cnode的前驱已经是一个排序的双向链表！BSTreeNode * convert2(BSTreeNode *cnode,BSTreeNode *orderedList){if(cnode == NULL){return NULL;}//先访问左子树if(cnode->m_pLeft != NULL){orderedList = convert2(cnode->m_pLeft,orderedList);}if(orderedList != NULL){//调整orderedList和cnode的关系cnode->m_pLeft = orderedList;orderedList->m_pRight = cnode;}//生成新的orderedListorderedList = cnode;//访问右子树if(cnode->m_pRight != NULL){orderedList = convert2(cnode->m_pRight,orderedList);}return orderedList;//返回的是最大节点的指针，这点和方法一不同(方法一也可以改成这样！)}//生成测试的二叉查找树BSTreeNode * create(){int data[]={10,6,14,4,8,12,16};int i;BSTreeNode * tree=NULL;for(i=0;i<sizeof(data)/sizeof(int);i++){BSTreeNode *fresh = (BSTreeNode*)malloc(sizeof(BSTreeNode));fresh->m_value = data[i];fresh->m_pLeft = fresh->m_pRight = NULL;tree = insert(tree,fresh);}return tree;}//测试方法一void test_convert(){BSTreeNode *p=NULL, *fptr=NULL;BSTreeNode *tree = create();tree = convert(tree,-1);p=tree;while(p != NULL){printf("%d,",p->m_value);fptr = p;p = p->m_pRight;free(fptr);}printf("\n");}//测试方法二void test_convert2(){BSTreeNode *p=NULL, *fptr=NULL;BSTreeNode *tree = create();printf("convert2:\n");tree = convert2(tree,NULL);p=tree;while(p != NULL){//从右向左遍历printf("%d,",p->m_value);fptr=p;p=p->m_pLeft;free(fptr);}printf("\n");}int main(){test_convert();test_convert2();return 0;}

题二：

#include<stdio.h>#include<stdlib.h>#define MAXSIZE 20typedef struct BSTreeNode{int m_value;struct BSTreeNode *m_pLeft;struct BSTreeNode *m_pRight;}BSTreeNode;//先得写个简单的int栈typedef struct stack{BSTreeNode* data[MAXSIZE];int top;}Stack;//-------- 栈的各种操作--------------------//int isEmpty(Stack *s){if(s->top == 0){return 1;}else{return 0;}}int push(Stack *s, BSTreeNode *node){if(s->top>=MAXSIZE){printf("error:栈已满\n");return -1;}s->data[s->top++] = node;return 1;}BSTreeNode * pop(Stack *s){if(s->top == 0){printf("error:栈内没有元素\n");return NULL;}return s->data[--(s->top)];}//------- 树的各种操作 --------------------------////生成二叉查找树BSTreeNode *insert(BSTreeNode *tree,BSTreeNode *n){if(tree == NULL){tree = n;}else if(n->m_value < tree->m_value){tree->m_pLeft = insert(tree->m_pLeft,n);}else{tree->m_pRight = insert(tree->m_pRight,n);}return tree;}//生成测试的二叉查找树BSTreeNode * create(){int data[]={10,6,14,4,8,12,16};int i;BSTreeNode * tree=NULL;for(i=0;i<sizeof(data)/sizeof(int);i++){BSTreeNode *fresh = (BSTreeNode*)malloc(sizeof(BSTreeNode));fresh->m_value = data[i];fresh->m_pLeft = fresh->m_pRight = NULL;tree = insert(tree,fresh);}return tree;}//非递归的将二叉查找树变为其镜像BSTreeNode * mirror(BSTreeNode *cnode){if(cnode == NULL){return NULL;}Stack s;s.top=0;//初始化栈指针BSTreeNode *p = cnode, *tmp=NULL;while(cnode!=NULL || !isEmpty(&s)){if(cnode){push(&s,cnode);cnode = cnode->m_pLeft;}else{cnode = pop(&s);//交换cnode的左右孩子tmp = cnode->m_pLeft;cnode->m_pLeft = cnode->m_pRight;cnode->m_pRight = tmp;cnode = cnode->m_pLeft;}}return p;}//递归版本BSTreeNode *re_mirror(BSTreeNode *cnode){BSTreeNode * tmp=NULL;if(cnode != NULL){//交换左右子树的指针tmp = cnode->m_pLeft;cnode->m_pLeft = cnode->m_pRight;cnode->m_pRight = tmp;//递归cnode->m_pLeft = re_mirror(cnode->m_pLeft);cnode->m_pRight = re_mirror(cnode->m_pRight);}return cnode;}void printTree(BSTreeNode *cnode){if(cnode != NULL){printTree(cnode->m_pLeft);printf("%d,",cnode->m_value);printTree(cnode->m_pRight);}}int main(){BSTreeNode *tree = create();tree = re_mirror(tree);BSTreeNode *p=NULL, *fptr=NULL;printTree(tree);printf("\n");return 0;}

小结：要想解决二叉树中的大多数问题一定要理解"递归"的思想！而且要知道非递归算法的写法！我自己对非递归算法都还有点不熟，汗！！

本文重点参考了 http://zhedahht.blog.163.com/ 的内容，深表感谢！！

我的代码都是用纯C语言实现的，而且分析写在了注释中。代码没有做过"优化"，主要是便于理解解题思想！

(全文完)