二叉树前序、中序、后序非递归遍历

来源：互联网发布：visio 数据流程图编辑：程序博客网时间：2024/04/28 06:27
#include <iostream>#include <stack>using namespace std;using std::stack;//链表节点template<class T>struct Node{    T data;    Node<T> *left;    Node<T> *right;    Node<T>(T t)    {        left=NULL;        right=NULL;        data=t;    }};//前序非递归template<class T>void preOrder(Node<T> *tree){    Node<T> *p=tree;    stack<Node<T> *> stk;    while(!stk.empty()||p)    {        if(p)        {            cout<<p->data<<endl;            stk.push(p);            p=p->left;        }        else        {            p=stk.top()->right;            stk.pop();        }    }}//中序非递归template<class T>void midOrder(Node<T> *tree){    Node<T> *p=tree;    if(!p)        return;    stack<Node<T> *> stk;    while(!stk.empty()||p)    {         while(p)        {        stk.push(p);        p=p->left;        }        p=stk.top();        cout<<p->data<<endl;        p=p->right;        stk.pop();    }}//后序非递归template<class T>void lastOrder(Node<T> *tree){    Node<T> *p=tree;    Node<T> *pre=p;    if(!p)        return;    stack<Node<T> *> stk;    while(!stk.empty()||p)    {        while(p)        {            stk.push(p);            p=p->left;        }        p=stk.top();        if(p->right&&p->right!=pre)        {            p=p->right;        }        else        {            cout<<p->data<<endl;            pre=p;            stk.pop();            p=NULL;        }    }}//后序非递归方案2，先按照根节点-右节点-左节点的顺序访问，然后reversetemplate<class T>void lastOrder(Node<T> *tree){Node<T> *p=tree;stack<Node<T> *> stk;vector<T> vec;while(p||!stk.empty()){    if(p)    {        vec.push_back(p->data);        stk.push(p);        p=p->right;    }    else    {        p=stk.top()->left;        stk.pop();    }}reverse(vec.begin(),vec.end());for(auto e:vec)    cout<<e<<" ";return 0;}//testint main(int argc, char* argv[]){  Node<int> *tree=new Node<int>(1);  Node<int> *p=tree;  p->left=new Node<int>(2);  p->right=new Node<int>(3);  p=p->left;  p->left=new Node<int>(4);  p->right=new Node<int>(5);  p=p->right;  p->right=new Node<int>(6);  p->left=new Node<int>(7);  lastOrder(tree);  return 0;}
0 0