平衡二叉树转化为双向链表

来源：互联网发布：学费软件编辑：程序博客网时间：2024/04/29 09:49

很容易想到递归，实现确实不是太容易，对本人来说。

平衡二叉树是有序的，要求链表也是有序。

代码：

#include<iostream>  //平衡二叉树转化为双向链表using namespace std;typedef struct tree{int data;struct tree *lchild;struct tree *rchild;}Tree,*pTree;void createBST(pTree &pHead){int temp;scanf("%d",&temp);if(temp){pHead = new Tree();pHead->data = temp;createBST(pHead->lchild);createBST(pHead->rchild);}else{pHead = NULL;areturn ;}}void print(pTree p){if(p){print(p->lchild);cout<<p->data<<" ";print(p->rchild);}else{  return ;}}void convertList(pTree root,pTree * pLastNode){if(root == NULL)return ;pTree cur = root;if(cur->lchild != NULL){convertList(root->lchild,pLastNode);}cur->lchild = *pLastNode;if(*pLastNode != NULL){(*pLastNode)->rchild = cur;}*pLastNode = cur;if(cur->rchild != NULL){convertList(cur->rchild,pLastNode);}}void printList(const pTree cur1){pTreecur = cur1;while(cur){cout<<cur->data<<" ";cur = cur->rchild;}}pTree convert(pTree root){if(NULL == root)return root;pTree  pLastNode = NULL;convertList(root,& pLastNode);pTree pHead = pLastNode;while(pHead != NULL && pHead ->lchild != NULL){pHead = pHead->lchild;}return pHead;}int main()  {  pTree pHead = NULL;createBST(pHead);cout<<"原平衡二叉树的中序遍历：";print(pHead);cout<<endl;cout<<"链表的从左到右序列：";pTree pHead1=convert(pHead);printList(pHead1);    return 0;  }

0 0