C ++程序执行给定二叉树的后顺序非递归遍历
如果二叉树是后置遍历的，则首先访问左子树，然后访问右子树，然后再访问根。这是一个C++程序，用于在没有递归的情况下遍历后序树。我们在这里通过使用堆栈来执行程序。
算法
对于后遍历：
Begin
Declare postorder_traversal(struct node*t,struct tree**top)
if(t==NULL) then
print “Empty Tree”.
Return.
Print “Postorder Data Using Stack :”.
Call push(t,top) function to insert values.
Declare a pointer store against tree structure.
Initialize struct tree*store=NULL.
while(t!=NULL)
store=*top;
if(store->v==0) then
if(t->r!=NULL) then
(store->v)++
push(t->r,top)
if(t->l!=NULL) then
(store->v)++
push(t->l,top)
if(store->v==0) then
cout<d
t=NULL
pop(top)
else
cout<d
t=NULL
pop(top)
if(*top!=NULL) then
t=(*top)->link
End示例
#include<iostream>
#include<stdlib.h>
using namespace std;
struct node {
int d;
struct node *l,*r;
};
struct tree {
int v;
struct node*link;
struct tree*n;
};
struct node*create_node(int);
struct node*create_node(int value) {
struct node*new_node=(struct node*)malloc(sizeof(struct node));
if(new_node!=NULL) {
new_node->d=value;
new_node->l=new_node->r=NULL;
return new_node;
} else {
printf("\n Memory overflow.");
return NULL;
}
}
void push(struct node*,struct tree*);
void push(struct node*node,struct tree**top) {
struct tree*new_node=(struct tree*)malloc(sizeof(struct tree));
if(new_node!=NULL) {
new_node->link=node;
new_node->n=*top;
new_node->v=0;
*top=new_node;
} else {
cout<<"\n Memory overflow.";
return ;
}
}
void pop(struct tree**);
void pop(struct tree**top) {
if(*top!=NULL) {
struct tree*remove=*top;
*top=(*top)->n;
remove->link=NULL;
remove->n=NULL;
remove=NULL;
}
}
void postorder_traversal(struct node*,struct tree**);
void postorder_traversal(struct node*t,struct tree**top) {
if(t==NULL) {
cout<<"\n Empty Tree";
return;
}
cout<<"\n Postorder Data Using Stack :";
push(t,top);
struct tree*store=NULL;
while(t!=NULL) {
store=*top;
if(store->v==0) {
if(t->r!=NULL) {
(store->v)++;
push(t->r,top);
}
if(t->l!=NULL) {
(store->v)++;
push(t->l,top);
}
if(store->v==0) {
cout<<t->d;
t=NULL;
pop(top);
}
}
else {
cout<<t->d;
t=NULL;
pop(top);
}
if(*top!=NULL)
t=(*top)->link;
}
}
int main(){
struct node*root=NULL;
struct tree*top=NULL;
root = create_node(20);
root->l = create_node(10);
root->r = create_node(30);
root->r->r = create_node(7);
root->l->l = create_node(25);
root->l->r = create_node(35);
root->l->r->r = create_node(40);
root->l->l->r = create_node(26);
postorder_traversal(root,&top);
return 0;
}输出结果
Postorder Data Using Stack :26 25 40 35 10 7 30 20