标签:style c class blog code java
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; }
数据结构-二叉搜索树与双向链表,布布扣,bubuko.com
标签:style c class blog code java
原文地址:http://www.cnblogs.com/wn19910213/p/3747719.html