二叉树的遍历

1.思路
先序遍历：遍历顺序 root、leftchild、rightchild
中序遍历：遍历顺序leftchild、root、rightchild
后序遍历：遍历顺序leftchild、rightchild、root

“先中后”指的是根结点root在遍历顺序中的相对位置。程序中分别用递归与非递归方法实现，递归方法较为简单，非递归方法采用堆栈stack进行辅助操作。

2.代码

#include<iostream>using namespace std;struct node {    int value;    node* lchild;    node* rchild;};//0表示空结点node* create() {    int value;    node* root;    cin >> value;    if(value == 0) {        root = NULL;    } else {        root = (node*)malloc(sizeof(node));        root->value = value;        cout << "please input (" << value << ") its leftchild node" << endl;        root->lchild = create();        cout << "please input (" << value << ") its rightchild node" << endl;        root->rchild = create();    }    return root;}//0表示空结点，init()不对树的结点赋值node* init() {    int value;    node* root;    cin >> value;    if(value == 0) {        root = NULL;        return root;    } else {        root = (node*)malloc(sizeof(node));        cout << "please input (" << value << ") its leftchild node" << endl;        root->lchild = init();        cout << "please input (" << value << ") its rightchild node" << endl;        root->rchild = init();        return root;    }}void visit(node* p) {    cout << p->value << " ";}//递归版先序遍历void preorder(node* p) {    if(p == NULL) {        return;    }    visit(p);    preorder(p->lchild);    preorder(p->rchild);}//非递归版先序遍历void preorder1(node* p) {    node* stack[100];    int ntop = 0;    node* cur = p;    while(cur != NULL || ntop > 0) {        while(cur != NULL) { //访问所以左孩子节点，并依次入栈            visit(cur);            stack[ntop++] = cur;            cur = cur->lchild;        }        if(ntop > 0) {            cur = stack[--ntop]->rchild;        }    }}void inorder(node* p) {    if(p == NULL) {        return;    }    inorder(p->lchild);    visit(p);    inorder(p->rchild);}void inorder1(node* p) {    node* stack[100];    int ntop = 0;    node* cur = p;    while(cur != NULL || ntop > 0) {        while(cur != NULL) {            stack[ntop++] = cur;            cur = cur->lchild;        }        if(ntop > 0) {            cur = stack[--ntop];            visit(cur);            cur = cur->rchild;        }    }}void posorder(node* p) {    if(p == NULL) {        return;    }    posorder(p->lchild);    posorder(p->rchild);    visit(p);}void posorder1(node* p) {    node* stack[100];    int ntop = 0;    node* cur = p;    node* lastvisit = NULL;    while(cur != NULL || ntop > 0) {        while(cur != NULL) {            stack[ntop++] = cur;            cur = cur->lchild;        }        cur = stack[ntop - 1];        if(cur->rchild == NULL || cur->rchild == lastvisit) {            visit(cur);            ntop--;            lastvisit = cur;            cur = NULL;        //cur = NULL 只是把cur清除，原二叉树的结点并没有变化！        } else {            cur = cur->rchild;        }    }}int main() {    cout << "please input the tree" << endl;    node* p = create();    cout << "preorder: ";    preorder1(p);    cout << endl;    cout << "inorder: ";    inorder1(p);    cout << endl;    cout << "posorder: ";    posorder1(p);    cout << endl;    return 0;}

3.运行结果