实现二叉树的前序/中序/后序非递归遍历——题集（十一）

非递归实现二叉树的前序/中序/后序遍历——题集（十一）

       今天分享一下非递归实现二叉树的前序/中序/后序遍历，以及用不含头结点的单链表存储集合A和B,求解集合A与B的差集，并将结果保存在集合A的单链表中。。

       实现二叉树的前序/中序/后序非递归遍历的源代码和运行示例。

源代码如下：

#include<iostream>using namespace std;#include<stack> //二叉树的前序/中序/后序非递归遍历struct TreeNode {    int val;    struct TreeNode *left;    struct TreeNode *right;    TreeNode(int x):val(x),left(NULL),right(NULL){    }}; //前序非递归遍历void PreTraNR(TreeNode* root){//借助栈cout<<"前序非递归遍历: ";if(root==NULL){cout<<endl;return;}TreeNode* cur=root;stack<TreeNode*> tmp;tmp.push(cur); while(!tmp.empty()){cur=tmp.top(); while(cur->left != NULL){cout<<cur->val<<" ";cur=cur->left;tmp.push(cur);}cout<<cur->val<<" "; while(cur->right==NULL){tmp.pop();if(tmp.empty()){cout<<endl;return;}cur=tmp.top();}tmp.pop();cur=cur->right;tmp.push(cur);} cout<<endl;return;} //中序非递归遍历void OrdTraNR(TreeNode* root){//借助栈cout<<"中序非递归遍历: ";if(root==NULL){cout<<endl;return;}TreeNode* cur=root;stack<TreeNode*> tmp;tmp.push(cur); while(!tmp.empty()){cur=tmp.top(); while(cur->left != NULL){cur=cur->left;tmp.push(cur);} cout<<cur->val<<" "; while(cur->right==NULL){tmp.pop();if(tmp.empty()){cout<<endl;return;}cur=tmp.top();cout<<cur->val<<" ";}tmp.pop();cur=cur->right;tmp.push(cur);} cout<<endl;return;} //后序非递归遍历void PostTraNR(TreeNode* root){//借助栈cout<<"后序非递归遍历: ";if(root==NULL){cout<<endl;return;}TreeNode* cur=root;TreeNode* prev=NULL;stack<TreeNode*> tmp;tmp.push(cur); while(!tmp.empty()){cur=tmp.top(); while(cur->left != NULL){//prev=cur;cur=cur->left;tmp.push(cur);} while(cur->right == prev || cur->right==NULL){cout<<cur->val<<" ";prev=cur;tmp.pop();if(tmp.empty()){cout<<endl;return;}cur=tmp.top();}cur=cur->right;tmp.push(cur);} cout<<endl;return;} void TestTree()//二叉树的前序/中序/后序非递归遍历{cout<<"二叉树的前序/中序/后序非递归遍历"<<endl<<endl;TreeNode * pRoot1=new TreeNode(1);TreeNode * pRoot2=new TreeNode(2);TreeNode * pRoot3=new TreeNode(3);TreeNode * pRoot4=new TreeNode(4);TreeNode * pRoot5=new TreeNode(5);TreeNode * pRoot6=new TreeNode(6);TreeNode * pRoot7=new TreeNode(7);TreeNode * pRoot8=new TreeNode(8);TreeNode * pRoot9=new TreeNode(9); pRoot1->left = pRoot2;pRoot1->right = pRoot3;pRoot2->left = pRoot4;pRoot2->right = pRoot5;pRoot3->left = pRoot6;pRoot3->right = pRoot7;pRoot4->left = pRoot8;PreTraNR(pRoot1);//前序非递归遍历OrdTraNR(pRoot1);//中序非递归遍历 PostTraNR(pRoot1);//后序非递归遍历} int main(){TestTree();//二叉树的前序/中序/后序非递归遍历 system("pause");return 0;}

运行结果：

 

       已知集合A和B的元素分别用不含头结点的单链表存储，函数difference()用于求解集合A与B的差集，并将结果保存在集合A的单链表中。例如，若集合A={5,10,20,15,25,30}，集合B={5,15,35,25}，完成计算后A={10,20,30}。

      链表结点的结构类型定义如下：

struct node{int elem;node* next;};

       用不含头结点的单链表存储集合A和B,求解集合A与B的差集，并将结果保存在集合A的单链表中。其函数void difference(node** LA , node* LB)的源代码和运行示例。

源代码如下：

#include<iostream>using namespace std;//用不含头结点的单链表存储集合A和B,求解集合A与B的差集，并将结果保存在集合A的单链表中。struct ListNode{int val;ListNode* next;ListNode(int _val):val(_val),next(NULL){}};void  DifSet(ListNode** LA, ListNode* LB ){if(LB == NULL || LA == NULL) return;ListNode* cur=*LA;while(LB){cur=*LA;ListNode* prev=NULL;while(cur){if(cur->val == LB->val){if(prev)prev->next=cur->next;else*LA=cur->next;delete cur;break;}prev=cur;cur=cur->next;}LB=LB->next;}return;}void TestSet(){//求解集合A与B的差集ListNode * lA1=new ListNode(1);ListNode * lA2=new ListNode(2);ListNode * lA3=new ListNode(3);ListNode * lA4=new ListNode(4);ListNode * lA5=new ListNode(5);ListNode * lA6=new ListNode(6);ListNode * lA7=new ListNode(7);ListNode * lA8=new ListNode(8);lA1->next = lA2;lA2->next = lA3;lA3->next = lA4;lA4->next = lA5;lA5->next = lA6;lA6->next = lA7;lA7->next = lA8;ListNode * lB1 = new ListNode(1);ListNode * lB2 = new ListNode(4);ListNode * lB3 = new ListNode(6);ListNode * lB4 = new ListNode(8);ListNode * lB5 = new ListNode(11);lB1->next = lB2;lB2->next = lB3;lB3->next = lB4;lB4->next = lB5;cout<<"求解不含头结点的单链表存储集合lA1与lB1的差集,"<<endl<<"并将结果保存在集合lA1的单链表中"<<endl<<endl;cout<<"打印集合lA1: ";Printf(lA1);cout<<"打印集合lB1: ";Printf(lB1);DifSet( &lA1, lB1);cout<<"DifSet( &lA1, lB1)->打印集合lA1与lB1的差集结果; ";Printf(lA1);}int main(){TestSet();//求解集合A与B的差集system("pause");return 0;}

运行结果：

 

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