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。
首先我们定义的二元查找树节点的数据结构如下：
struct BSTreeNode
{
int m_nValue; // value of node
BSTreeNode *m_pLeft; // left child of node
BSTreeNode *m_pRight; // right child of node
};
#include <stdio.h>
#include <iostream.h>
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 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;
}
}
}
// 遍历二元查找树中序
void ergodicBSTree(BSTreeNode * pCurrent)
{
if (NULL == pCurrent)
{
return;
}
if (NULL != pCurrent->m_pLeft)
{
ergodicBSTree(pCurrent->m_pLeft);
}
// 节点接到链表尾部
convertToDoubleList(pCurrent);
// 右子树为空
if (NULL != pCurrent->m_pRight)
{
ergodicBSTree(pCurrent->m_pRight);
}
}
// 二叉树转换成list
void convertToDoubleList(BSTreeNode * pCurrent)
{
pCurrent->m_pLeft = pListIndex;
if (NULL != pListIndex)
{
pListIndex->m_pRight = pCurrent;
}
else
{
pHead = pCurrent;
}
pListIndex = pCurrent;
cout<<pCurrent->m_nValue<<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);
return 0;
}
