剑指offer之二叉搜索树和双向链表

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

4
           2         6
        1     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