把二元查找树转变成排序的双向链表（树）

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题目：

输入一颗二元查找树，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节点，只调整指针的指向。

例如 10

/ \

6 14

/ \ / \

4 8 12 16

转换成双向链表： 4=6=8=10=12=14=16

递归方法

递归函数返回二元查找树转换后的双向链表的首尾指针，先转换节点的两棵子树，然后再将两棵子树转换得到的两个双向链表通过该节点连接起来。

#include<iostream>using namespace std;struct BSTreeNode{    int m_nValue;    BSTreeNode *m_pLeft;    BSTreeNode *m_pRight;};typedef pair<BSTreeNode*, BSTreeNode*> HeadandTail;HeadandTail BSTreeToDlist(BSTreeNode *root){    if(root == NULL)        return HeadandTail(NULL,NULL);    BSTreeNode *headPtr = root;    BSTreeNode *endPtr = root;    if(root->m_pLeft != NULL)    {        HeadandTail lResult = BSTreeToDlist(root->m_pLeft);        headPtr = lResult.first;        lResult.second->m_pRight = root;        root->m_pLeft = lResult.second;    }    if(root->m_pRight != NULL)    {        HeadandTail rResult = BSTreeToDlist(root->m_pRight);        endPtr = rResult.second;        rResult.first->m_pLeft = root;        root->m_pRight = rResult.first;    }    return HeadandTail(headPtr, endPtr);}int main(){    /* 测试树            10                         /\                        /  \                       6   14    */    BSTreeNode *root = new BSTreeNode;    root->m_nValue = 10;    BSTreeNode *left = new BSTreeNode;    left->m_nValue = 6;    left->m_pLeft = NULL;    left->m_pRight = NULL;    BSTreeNode *right = new BSTreeNode;    right->m_nValue = 14;    right->m_pLeft = NULL;    right->m_pRight = NULL;    root->m_pLeft = left;    root->m_pRight = right;    HeadandTail result = BSTreeToDlist(root);    BSTreeNode *head = result.first;    BSTreeNode *tail = result.second;    while(head != NULL)    {        cout<<head->m_nValue<<" ";        head = head->m_pRight;    }    cout<<endl;    while(tail != NULL)    {        cout<<tail->m_nValue<<" ";        tail = tail->m_pLeft;    }    cout<<endl;    return 0;}

非递归方法

该问题的实质是中序遍历二元树，将当前遍历的节点和前一次遍历的节点连接起来即可。

源码待补充

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