面试100题：1 把二元查找树转变成排序的双向链表

最新推荐文章于 2026-01-05 16:50:18 发布

原创最新推荐文章于 2026-01-05 16:50:18 发布 · 726 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#面试 #null #delete #system #list

本文介绍如何将二元查找树转换为排序的双向链表，通过递归方式实现，不创建新节点，仅调整节点间的指针连接。讨论了内存管理和打印功能的问题。

1.把二元查找树转变成排序的双向链表
题目：
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。

现存问题：
1 造成内存泄露，因为只有new，没有delete
2 printNode（）函数打印结果与预期不符

#include <stdio.h> #include <iostream> #include <iomanip> using namespace std; struct BSTreeNode { int m_nValue; // value of node BSTreeNode *m_pLeft; // left child of node BSTreeNode *m_pRight; // right child of node }; typedef BSTreeNode DoubleList; DoubleList * pHead; DoubleList * pListIndex; void convertToDoubleList(BSTreeNode * pCurrent); // 创建二元查找树 void printNode(BSTreeNode * pCurrent); void addBSTreeNode(BSTreeNode *&pCurrent, int value) { if (NULL == pCurrent) { BSTreeNode * pBSTree = new BSTreeNode(); pBSTree->m_pLeft = NULL; pBSTree->m_pRight = NULL; pBSTree->m_nValue = value; pCurrent = pBSTree; } else { if ((pCurrent->m_nValue)>value) { addBSTreeNode(pCurrent->m_pLeft, value); } else if ((pCurrent->m_nValue)<value) { addBSTreeNode(pCurrent->m_pRight, value); } else { //cout<<"重复加入节点"<<endl; } } } void ergodicBSTree(BSTreeNode * pCurrent)// 遍历二元查找树中序 { if (NULL == pCurrent) { return; } if (NULL != pCurrent->m_pLeft) { printNode(pCurrent); ergodicBSTree(pCurrent->m_pLeft); } convertToDoubleList(pCurrent);// 节点接到链表尾部 if (NULL != pCurrent->m_pRight)// 右子树为空 { printNode(pCurrent); ergodicBSTree(pCurrent->m_pRight); } } void printNode(BSTreeNode * pCurrent) { if(NULL==pCurrent->m_pLeft) cout<<"Left == NULL"<<" , "; else cout<<"Left == "<<pCurrent->m_pLeft->m_nValue<<" , "; cout<<"Current == "<<pCurrent->m_nValue<<" , "; if(NULL==pCurrent->m_pRight) cout<<"Right == NULL"<<endl; else cout<<"Right == "<<pCurrent->m_pRight->m_nValue<<endl; } void convertToDoubleList(BSTreeNode * pCurrent)// 二叉树转换成list { pCurrent->m_pLeft = pListIndex; if (NULL != pListIndex) { pListIndex->m_pRight = pCurrent; } else { pHead = pCurrent; } pListIndex = pCurrent; //cout<<pCurrent->m_nValue<<endl; } void printDoubleList(DoubleList *pHead) { pListIndex=pHead; while(NULL!=pListIndex->m_pRight) { cout<<pListIndex->m_nValue<<" "; pListIndex=pListIndex->m_pRight; } cout<<endl; } int main() { BSTreeNode * pRoot = NULL; pListIndex = NULL; pHead = NULL; addBSTreeNode(pRoot, 10); addBSTreeNode(pRoot, 4); addBSTreeNode(pRoot, 6); addBSTreeNode(pRoot, 8); addBSTreeNode(pRoot, 12); addBSTreeNode(pRoot, 14); addBSTreeNode(pRoot, 15); addBSTreeNode(pRoot, 16); ergodicBSTree(pRoot); printDoubleList(pHead); system("pause");return 0; }

打印结果：

Left == 4 , Current == 10 , Right == 12
Left == NULL , Current == 4 , Right == 6
Left == 4 , Current == 6 , Right == 8
Left == 8 , Current == 10 , Right == 12
Left == 10 , Current == 12 , Right == 14
Left == 12 , Current == 14 , Right == 15
Left == 14 , Current == 15 , Right == 16
4 6 8 10 12 14 15