pat-a 1043. Is It a Binary Search Tree (25)

很明显这个题我冗余代码写得有点多，昨天想使用指针来建树，一直出现指针指向非法空间问题没做出来。改成数组就对了。指针使用不是很熟悉。同学做这个题用递归建树，如果递归过程中建不了二叉树就直接输出NO，递归我写起来也是很容易越界。

这个题有相同的编号因此不能使用STL来代替建树

这个题题意就是判断所给数列是不是搜索二叉树或者镜像的搜索二叉树，输出两种树分别对应的后序遍历就行了（镜像树先右后左）

#include<cstdio>
#include<iostream>
#include<map>
#include<vector>
using namespace std;
struct node{
int num;
int l,r;
};
node tree[1010];
int nu[1010];
int temp[1010];
int te[1010];
int p=1,k=0,w=0,v=0;
void creat(int t){
int root=0;
tree[p].num=t;
do{
if(tree[root].num>t){
if(tree[root].l==-1){
tree[root].l=p;
break;
}
else root=tree[root].l;
}
else{
if(tree[root].r==-1){
tree[root].r=p;
break;
}
else root=tree[root].r;
}
}while(1);
p++;
}
void preorder(int root){
temp[k++]=tree[root].num;
if(tree[root].l!=-1) preorder(tree[root].l);
if(tree[root].r!=-1) preorder(tree[root].r);
}
void preorder1(int root){
te[v++]=tree[root].num;
if(tree[root].r!=-1) preorder1(tree[root].r);
if(tree[root].l!=-1) preorder1(tree[root].l);
}
void pastorder(int root){
if(tree[root].l!=-1) pastorder(tree[root].l);
if(tree[root].r!=-1) pastorder(tree[root].r);
if(w) printf(" %d",tree[root].num);
else{
w=1;
printf("%d",tree[root].num);
}
}
void pastorder1(int root){
if(tree[root].r!=-1) pastorder1(tree[root].r);
if(tree[root].l!=-1) pastorder1(tree[root].l);
if(w) printf(" %d",tree[root].num);
else{
w=1;
printf("%d",tree[root].num);
}
}
int main(){
int n,root=0,flag=1,flag1=1;
scanf("%d",&n);
for(int i=0;i<n;++i){
tree[i].num=tree[i].l=tree[i].r=-1;
}
for(int i=0;i<n;++i){
scanf("%d",&nu[i]);
if(i==0) tree[0].num=nu[i];
else creat(nu[i]);
}
preorder(0);
for(int i=0;i<n;++i){
if(nu[i]!=temp[i]){
flag=0;
break;
}
}
if(flag==0){
flag=1;
preorder1(0);
for(int i=0;i<n;++i){
if(nu[i]!=te[i]){
flag=0;
break;
}
}
if(flag){
printf("YES\n");
pastorder1(0);
}
else
printf("NO\n");
}
else{
printf("YES\n");
pastorder(0);
}
return 0;
}

A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties:

The left subtree of a node contains only nodes with keys less than the node's key.

The right subtree of a node contains only nodes with keys greater than or equal to the node's key.

Both the left and right subtrees must also be binary search trees.

If we swap the left and right subtrees of every node, then the resulting tree is called the Mirror Image of a BST.

Now given a sequence of integer keys, you are supposed to tell if it is the preorder traversal sequence of a BST or the mirror image of a BST.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (<=1000). Then N integer keys are given in the next line. All the numbers in a line are separated by a space.

Output Specification:

For each test case, first print in a line "YES" if the sequence is the preorder traversal sequence of a BST or the mirror image of a BST, or "NO" if not. Then if the answer is "YES", print in the next line the postorder traversal sequence of that tree. All
the numbers in a line must be separated by a space, and there must be no extra space at the end of the line.
Sample Input 1:

7
8 6 5 7 10 8 11

Sample Output 1:

YES
5 7 6 8 11 10 8

Sample Input 2:

7
8 10 11 8 6 7 5

Sample Output 2:

YES
11 8 10 7 5 6 8

Sample Input 3:

7
8 6 8 5 10 9 11

Sample Output 3:

NO