1 问题

比如我们搜索二叉树如下，我们需要变成双向链表
 

 

 

 

 

 

 

2 分析

我们知道这个变成双向链接的时候是按照树的中序遍历打印的，我们只需要在中序遍历打印的时候操作该节点，我们可以用临时变量保存这个节点，同时我们也需要单独增加一个链表节点变量，我们需要保证这个节点的左边指向是该链表节点，然后该链表节点的右指向是这个节点，然后我们再把这个节点赋值给这个链表节点，就这样一直移动下去即可。

 

 

 

 

 

3 代码实现

我这里以下面的搜索二叉树进行操作的

42         61     3   5     7     

#include <stdio.h>
#include <stdlib.h>typedef struct Tree
{int value;struct Tree* left;struct Tree* right;
} Tree;/** 把搜索二叉树转成双向链表，我们按照中序遍历*/
void convertNode(Tree* node, Tree** lastNodeList)
{if (node == NULL)return;Tree* pCurrent = node;if (pCurrent->left != NULL){convertNode(pCurrent->left, lastNodeList);}//这里就要进行我们每个节点的前后正确的指向了pCurrent->left = *lastNodeList;if (*lastNodeList != NULL){(*lastNodeList)->right = pCurrent;}*lastNodeList = pCurrent;if (pCurrent->right != NULL){convertNode(pCurrent->right, lastNodeList);}
}/** 把翻转好的双向链表尾巴节点移动到链表头*/
Tree* convert(Tree* node, Tree* lastNodeList)
{if (node == NULL || lastNodeList == NULL){printf("node is NULL or lastNodeList is NULL\n");return NULL;}Tree* last = NULL;convertNode(node, &lastNodeList);//因为这个时候lastNodeList已经到了双向链表尾巴，我们需要移动到链表头Tree* headNodeList = lastNodeList;while (headNodeList != NULL && headNodeList->left != NULL){headNodeList = headNodeList -> left;} return headNodeList;
}/** 双向链表从左到右打印*/
void printRightList(Tree* headNodeList)
{if (headNodeList == NULL){printf("headNodeList is NULL\n");return;}printf("we will print list from left to right\n");Tree* pCurrent = headNodeList;while (pCurrent != NULL){printf("value is %d\n", pCurrent->value);pCurrent = pCurrent->right;}
}/** 双向链表从右到左打印*/
void printLeftList(Tree* headNodeList)
{if (headNodeList == NULL){printf("headNodeList is NULL\n");return;}printf("we will print list from right to left\n");Tree* pCurrent = headNodeList;//先把链表头结点移动为链表的尾巴while (pCurrent->right != NULL){//printf("value is %d\n", pCurrent->value);pCurrent = pCurrent->right;}//pCurrent = pCurrent->left;while (pCurrent != NULL){printf("value is %d\n", pCurrent->value);pCurrent = pCurrent->left;} 
}int main(void) 
{Tree *node1 , *node2 , *node3, *node4, *node5, *node6, *node7;node1 = (Tree *)malloc(sizeof(Tree));node2 = (Tree *)malloc(sizeof(Tree));node3 = (Tree *)malloc(sizeof(Tree));node4 = (Tree *)malloc(sizeof(Tree));node5 = (Tree *)malloc(sizeof(Tree));node6 = (Tree *)malloc(sizeof(Tree));node7 = (Tree *)malloc(sizeof(Tree)); node1->value = 4;node2->value = 2;node3->value = 6;node4->value = 1;node5->value = 3;node6->value = 5;node7->value = 7;node1->left = node2;node1->right = node3;node2->left = node4;node2->right = node5;node3->left = node6;node3->right = node7;node4->left = NULL;node4->right = NULL;node5->left = NULL;node5->right = NULL;node6->left = NULL;node6->right = NULL;node7->left = NULL;node7->right = NULL;Tree* list = (Tree *)malloc(sizeof(Tree));if (!list){printf("malloc list fail\n");return -1;}Tree* firstNodeList = NULL;//convertNode(node1, &list);firstNodeList = convert(node1, list);if (firstNodeList == NULL){printf("firstNodeList is NULL\n");return -1;}printRightList(firstNodeList);printLeftList(firstNodeList);return 0;
}

 

 

 

 

4 运行结果

we will print list from left to right
value is 0
value is 1
value is 2
value is 3
value is 4
value is 5
value is 6
value is 7
we will print list from right to left
value is 7
value is 6
value is 5
value is 4
value is 3
value is 2
value is 1
value is 0