面试题 27(*)，二叉搜索树与双向链表(二叉搜索树的中序遍历就是有序序列，二叉树和双向链表都是每个结点两个指针)

 

思路：二叉搜索树的中序遍历就是有序序列，递归实现。递归终止条件是遇到NULL结点。

但是和打印结点不同，我们需要知道当前结点的前一次递归处理了什么结点，前一次递归处理的结点就是当前递归处理结点的prev结点，前一次递归结点的next就是当前结点。为此我们用了一个公共变量BinaryTreeNode  **temp，这个指针所指向的地址，专门用来存放当前被处理的结点的地址。

这样，每次递归函数中，先处理左子树，然后从temp中取出上次处理的结点，把当前结点的left指针指向temp，然后将temp的right指针指向当前结点。然后把temp赋值为当前结点，再调用递归函数处理右子树。

边界条件：空树。

代码，核心在treeTransform函数里。

#include "stdafx.h"

struct BinaryTreeNode 
{
    int                    m_nValue; 
    BinaryTreeNode*        m_pLeft;  
    BinaryTreeNode*        m_pRight; 
};

BinaryTreeNode* CreateBinaryTreeNode(int value)
{
    BinaryTreeNode* pNode = new BinaryTreeNode();
    pNode->m_nValue = value;
    pNode->m_pLeft = NULL;
    pNode->m_pRight = NULL;

    return pNode;
}

void ConnectTreeNodes(BinaryTreeNode* pParent, BinaryTreeNode* pLeft, BinaryTreeNode* pRight)
{
    if(pParent != NULL)
    {
        pParent->m_pLeft = pLeft;
        pParent->m_pRight = pRight;
    }
}

void PrintTreeNode(BinaryTreeNode* pNode)
{
    if(pNode != NULL)
    {
        printf("value of this node is: %d\n", pNode->m_nValue);

        if(pNode->m_pLeft != NULL)
            printf("value of its left child is: %d.\n", pNode->m_pLeft->m_nValue);
        else
            printf("left child is null.\n");

        if(pNode->m_pRight != NULL)
            printf("value of its right child is: %d.\n", pNode->m_pRight->m_nValue);
        else
            printf("right child is null.\n");
    }
    else
    {
        printf("this node is null.\n");
    }

    printf("\n");
}

void PrintTree(BinaryTreeNode* pRoot)
{
    PrintTreeNode(pRoot);

    if(pRoot != NULL)
    {
        if(pRoot->m_pLeft != NULL)
            PrintTree(pRoot->m_pLeft);

        if(pRoot->m_pRight != NULL)
            PrintTree(pRoot->m_pRight);
    }
}

void DestroyTree(BinaryTreeNode* pRoot)
{
    if(pRoot != NULL)
    {
        BinaryTreeNode* pLeft = pRoot->m_pLeft;
        BinaryTreeNode* pRight = pRoot->m_pRight;

        delete pRoot;
        pRoot = NULL;

        DestroyTree(pLeft);
        DestroyTree(pRight);
    }
}


void *treeTransform(BinaryTreeNode *node, BinaryTreeNode **temp){
    if(NULL == node)
        return NULL;
        
    treeTransform(node -> m_pLeft, temp);
    node -> m_pLeft = *temp;
    if(*temp != NULL)
        (*temp) -> m_pRight = node;
    *temp = node;
    treeTransform(node -> m_pRight, temp);
}

BinaryTreeNode* Convert(BinaryTreeNode* pRootOfTree)
{
    BinaryTreeNode *pLastNodeInList = NULL;
    treeTransform(pRootOfTree, &pLastNodeInList);

    // pLastNodeInList指向双向链表的尾结点，
    // 我们需要返回头结点
    BinaryTreeNode *pHeadOfList = pLastNodeInList;
    while(pHeadOfList != NULL && pHeadOfList->m_pLeft != NULL)
        pHeadOfList = pHeadOfList->m_pLeft;

    return pHeadOfList;
}

// ====================测试代码====================
void PrintDoubleLinkedList(BinaryTreeNode* pHeadOfList)
{
    BinaryTreeNode* pNode = pHeadOfList;

    printf("The nodes from left to right are:\n");
    while(pNode != NULL)
    {
        printf("%d\t", pNode->m_nValue);

        if(pNode->m_pRight == NULL)
            break;
        pNode = pNode->m_pRight;
    }

    printf("\nThe nodes from right to left are:\n");
    while(pNode != NULL)
    {
        printf("%d\t", pNode->m_nValue);

        if(pNode->m_pLeft == NULL)
            break;
        pNode = pNode->m_pLeft;
    }

    printf("\n");
}

void DestroyList(BinaryTreeNode* pHeadOfList)
{
    BinaryTreeNode* pNode = pHeadOfList;
    while(pNode != NULL)
    {
        BinaryTreeNode* pNext = pNode->m_pRight;

        delete pNode;
        pNode = pNext;
    }
}

void Test(char* testName, BinaryTreeNode* pRootOfTree)
{
    if(testName != NULL)
        printf("%s begins:\n", testName);

    PrintTree(pRootOfTree);

    BinaryTreeNode* pHeadOfList = Convert(pRootOfTree);

    PrintDoubleLinkedList(pHeadOfList);
}

//            10
//         /      \
//        6        14
//       /\        /\
//      4  8     12  16
void Test1()
{
    BinaryTreeNode* pNode10 = CreateBinaryTreeNode(10);
    BinaryTreeNode* pNode6 = CreateBinaryTreeNode(6);
    BinaryTreeNode* pNode14 = CreateBinaryTreeNode(14);
    BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
    BinaryTreeNode* pNode8 = CreateBinaryTreeNode(8);
    BinaryTreeNode* pNode12 = CreateBinaryTreeNode(12);
    BinaryTreeNode* pNode16 = CreateBinaryTreeNode(16);

    ConnectTreeNodes(pNode10, pNode6, pNode14);
    ConnectTreeNodes(pNode6, pNode4, pNode8);
    ConnectTreeNodes(pNode14, pNode12, pNode16);

    Test("Test1", pNode10);

    DestroyList(pNode4);
}

//               5
//              /
//             4
//            /
//           3
//          /
//         2
//        /
//       1
void Test2()
{
    BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
    BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
    BinaryTreeNode* pNode3 = CreateBinaryTreeNode(3);
    BinaryTreeNode* pNode2 = CreateBinaryTreeNode(2);
    BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1);

    ConnectTreeNodes(pNode5, pNode4, NULL);
    ConnectTreeNodes(pNode4, pNode3, NULL);
    ConnectTreeNodes(pNode3, pNode2, NULL);
    ConnectTreeNodes(pNode2, pNode1, NULL);

    Test("Test2", pNode5);

    DestroyList(pNode1);
}

// 1
//  \
//   2
//    \
//     3
//      \
//       4
//        \
//         5
void Test3()
{
    BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1);
    BinaryTreeNode* pNode2 = CreateBinaryTreeNode(2);
    BinaryTreeNode* pNode3 = CreateBinaryTreeNode(3);
    BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
    BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);

    ConnectTreeNodes(pNode1, NULL, pNode2);
    ConnectTreeNodes(pNode2, NULL, pNode3);
    ConnectTreeNodes(pNode3, NULL, pNode4);
    ConnectTreeNodes(pNode4, NULL, pNode5);

    Test("Test3", pNode1);

    DestroyList(pNode1);
}

// 树中只有1个结点
void Test4()
{
    BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1);
    Test("Test4", pNode1);

    DestroyList(pNode1);
}

// 树中没有结点
void Test5()
{
    Test("Test5", NULL);
}

int main()
{
    Test1();
    Test2();
    Test3();
    Test4();
    Test5();

    return 0;
}

 

书上代码基本上一样，但是多了几个非NULL判断，我没加也通过了，因为递归函数一开始就有判断NULL，应该不需要再额外判断了。

#include "stdafx.h"
#include "..\Utilities\BinaryTree.h"

void ConvertNode(BinaryTreeNode* pNode, BinaryTreeNode** pLastNodeInList);

BinaryTreeNode* Convert(BinaryTreeNode* pRootOfTree)
{
    BinaryTreeNode *pLastNodeInList = NULL;
    ConvertNode(pRootOfTree, &pLastNodeInList);

    // pLastNodeInList指向双向链表的尾结点，
    // 我们需要返回头结点
    BinaryTreeNode *pHeadOfList = pLastNodeInList;
    while(pHeadOfList != NULL && pHeadOfList->m_pLeft != NULL)
        pHeadOfList = pHeadOfList->m_pLeft;

    return pHeadOfList;
}

void ConvertNode(BinaryTreeNode* pNode, BinaryTreeNode** pLastNodeInList)
{
    if(pNode == NULL)
        return;

    BinaryTreeNode *pCurrent = pNode;

    if (pCurrent->m_pLeft != NULL)
        ConvertNode(pCurrent->m_pLeft, pLastNodeInList);

    pCurrent->m_pLeft = *pLastNodeInList; 
    if(*pLastNodeInList != NULL)
        (*pLastNodeInList)->m_pRight = pCurrent;

    *pLastNodeInList = pCurrent;

    if (pCurrent->m_pRight != NULL)
        ConvertNode(pCurrent->m_pRight, pLastNodeInList);
}