数据结构-二叉搜索树与双向链表

题目：输入一颗二叉搜索树，将该二叉搜索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，只能调整树中节点指针的指向。

分析：首先不能创建新的结构，只能在树的前提下进行改变指针的指向。又由于是二叉搜索树，可以通过画图分析可知，二叉搜索树的左子树小于根节点小于右子树，可以发现是个递归过程也是一个中序遍历。所以只需要在中序那块进行指针调整。

/*
剑指offer面试题27
*/
#include <iostream>

using namespace std;

struct BinaryTree{
    int data;
    BinaryTree* lchild;
    BinaryTree* rchild;
};

void Convert(BinaryTree* root,BinaryTree** pList){

    if(root == NULL){
        return;
    }

    BinaryTree* pTree = root;

    if(pTree->lchild){
        Convert(pTree->lchild,pList);
    }

    // 双链表
    pTree->lchild = *pList;
    if(*pList != NULL){
        (*pList)->rchild = pTree;
    }
    *pList = pTree;

    if(pTree->rchild){
        Convert(pTree->rchild,pList);
    }

}

BinaryTree* Convert(BinaryTree* root){
    BinaryTree* pList = NULL;
    Convert(root,&pList);

    BinaryTree* pListHead = pList;

    while(pListHead->lchild != NULL){
        pListHead = pListHead->lchild;
    }
    return pListHead;
}

BinaryTree* Create(){
    BinaryTree* root = new BinaryTree;
    root->data = 10;
    BinaryTree* lchild = new BinaryTree;
    lchild->data = 6;
    BinaryTree* rchild = new BinaryTree;
    rchild->data = 14;
    root->lchild = lchild;
    root->rchild = rchild;

    BinaryTree* lchild1 = new BinaryTree;
    lchild1->data = 4;
    BinaryTree* rchild1 = new BinaryTree;
    rchild1->data = 8;
    lchild->lchild = lchild1;
    lchild->rchild = rchild1;

    BinaryTree* lchild2 = new BinaryTree;
    lchild2->data = 12;
    BinaryTree* rchild2 = new BinaryTree;
    rchild2->data = 16;
    rchild->lchild = lchild2;
    rchild->rchild = rchild2;

    lchild1->lchild = NULL;
    lchild1->rchild = NULL;
    rchild1->lchild = NULL;
    rchild1->rchild = NULL;

    lchild2->lchild = NULL;
    lchild2->rchild = NULL;
    rchild2->lchild = NULL;
    rchild2->rchild = NULL;

    return root;
}

int main(){
    BinaryTree* root = Create();
    BinaryTree* pList = Convert(root);

    while(pList->rchild != NULL){
        cout << pList->data << " ";
        pList = pList->rchild;
    }
    cout << endl;
    while(pList->lchild != NULL){
        cout << pList->data << " ";
        pList = pList->lchild;
    }

    return 0;
}