HDU 1710二叉树的前序和中序遍历求后序遍历（结构体+指针）

Binary Tree Traversals

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 5238    Accepted Submission(s): 2417


[align=left]Problem Description[/align]
A binary tree is a finite set of vertices that is either empty or consists of a root r and two disjoint binary trees called the left and right subtrees. There are three most important ways in which the vertices of a binary tree can
be systematically traversed or ordered. They are preorder, inorder and postorder. Let T be a binary tree with root r and subtrees T1,T2.

In a preorder traversal of the vertices of T, we visit the root r followed by visiting the vertices of T1 in preorder, then the vertices of T2 in preorder.

In an inorder traversal of the vertices of T, we visit the vertices of T1 in inorder, then the root r, followed by the vertices of T2 in inorder.

In a postorder traversal of the vertices of T, we visit the vertices of T1 in postorder, then the vertices of T2 in postorder and finally we visit r.

Now you are given the preorder sequence and inorder sequence of a certain binary tree. Try to find out its postorder sequence.



 

[align=left]Input[/align]
The input contains several test cases. The first line of each test case contains a single integer n (1<=n<=1000), the number of vertices of the binary tree. Followed by two lines, respectively indicating the preorder sequence and
inorder sequence. You can assume they are always correspond to a exclusive binary tree.

 

[align=left]Output[/align]
For each test case print a single line specifying the corresponding postorder sequence.

 

[align=left]Sample Input[/align]

9
1 2 4 7 3 5 8 9 6
4 7 2 1 8 5 9 3 6

 

[align=left]Sample Output[/align]

7 4 2 8 9 5 6 3 1

题目要求：给你有n个结点的二叉树，接下来给你这个二叉树的前序遍历和中序遍历，让你输出他的后序遍历

思路很清晰：递归的套路嘛，

前序的第一个数x为当前根，在中序中找到这个数x，

中序x的左边就是左子树的中序遍历并统计数目为w1个，中序x右面就是右子树的中序遍历并统计数目为w2个。

前序除了第一个结点之后的w1个为左子树前序遍历，接下来的w2个为右子树前序遍历

然后先建树，然后后序遍历一遍，指针用的不6，写了好久。

#include<cstring>
#include<iostream>
#include<cstdio>
#include<conio.h>
using
4000
namespace std;
const int maxn = 1005 ;
struct node {
int date;
node *l;
node *r;
node(){
date = 0;
l=NULL;
r=NULL;
}
};
void create(node *t,int str1[],int len1,int str2[],int len2) {
if(len1==0) {
t = NULL ;
return ;
}
int index = 1;
for(int i=1; i<=len2; i++) {
if(str2[i]==str1[1]) {
index = i;
break ;
}
}
int ltemp[maxn];
int ltemplen = index-1;
int rtemp[maxn];
int rtemplen = len2-index;
for(int i=1; i<=ltemplen;i++)ltemp[i]=str2[i];
int c=1;
for(int i=index+1; i<=len2;i++)rtemp[c++]=str2[i];
int temp1[maxn];
int temp2[maxn];
c=1;
for(int i=2; i<=ltemplen+1;i++)temp1[c++] = str1[i];
c=1;
for(int i=ltemplen+2; i<=len1;i++)temp2[c++] = str1[i];
t->date = str1[1];
if(index-1!=0){
t->l=new node();
create(t->l,temp1,index-1,ltemp,index-1);
}
if(len2-index!=0){
t->r=new node();
create(t->r,temp2,len2-index,rtemp,len2-index);
}
return ;
}
int w=0;
int ans[maxn] ;
void lastorder(node *t) {
if(t!=NULL){
lastorder(t->l);
lastorder(t->r);
ans[w++]=t->date;
}
}
int main() {
int str1[maxn] ;
int str2[maxn] ;
int n ;
//freopen("in.txt","r",stdin);
while(~scanf("%d",&n)) {
w=0;
node *root = new node();
for(int i=1; i<=n; i++) {
scanf("%d",&str1[i]);
}
for(int i=1; i<=n; i++) {
scanf("%d",&str2[i]);
}
int len1=n;
int len2=n;
create(root,str1,len1,str2,len2);
lastorder(root);
for(int i=0;i<w-1;i++){
printf("%d ",ans[i]);
}printf("%d\n",ans[w-1]);
}
return 0;
}