//题目 通过后续遍历 中序遍历 得出一棵树 ,然后按树的层次遍历打印

PS:以前对于这种用指针的题目是比较头痛的,现在做了一些链表操作后,感觉也不难

先通过后续中序建一棵树,然后通过BFS遍历这棵树

提供测试样例

4
4 1 3 2
2 3 1 4
10
10 7 6 9 8 5 4 1 3 2
7 10 6 2 5 9 8 3 1 4

//题目 通过后续遍历 中序遍历 得出一棵树 ,然后按树的层次遍历打印
#include<stdio.h>
#include<iostream>
#include<queue>
using namespace std; int back[];
int mid[];
int n; struct data{
int v;
int no;
data *left,*right;
}; data *head=new data;
data *rhead=head;
void build(){
int i,j;
head->v=back[n];
head->left=NULL;
head->right=NULL; int root=n,rj;
for(j=;j<=n;j++){
if(back[n]==mid[j]){
head->no=j;break;
}
}
for(i=n-;i>=;i--){
for(j=;j<=n;j++){
if(back[i]==mid[j]){
rj=j;break;
}
}
head=rhead;
while(){
if((rj < head->no)&&head->left!=NULL){
head=head->left;continue;
}
if((rj > head->no)&&head->right!=NULL){
head=head->right;continue;
}break;
}
if(rj < head->no){
head->left=new data;
head=head->left;
head->v=back[i];
head->no=rj;
head->left=NULL;
head->right=NULL;
}else{
head->right=new data;
head=head->right;
head->v=back[i];
head->no=rj;
head->left=NULL;
head->right=NULL;
}
} } void bfs(){
data *first,*second;
first=rhead;
queue<data *>qq;
qq.push(first); int ok=;
while(!qq.empty()){ if(ok==){
ok=;
printf("%d",qq.front()->v);
}else{
printf(" %d",qq.front()->v);
}
second=qq.front();
qq.pop();
if(second->left!=NULL){
qq.push(second->left);
}
if(second->right!=NULL){
qq.push(second->right);
}
}
printf("\n");
} int main()
{
while(scanf("%d",&n)!=EOF){
int i;
head=new data;
rhead=head;
for(i=;i<=n;i++){
scanf("%d",&back[i]);
}
for(i=;i<=n;i++){
scanf("%d",&mid[i]);
}
build();
bfs();
} return ;
}

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