A1043 Is It a Binary Search Tree （25 分)

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

算法思路简单描述：先不管是不是BST，先根据输入序列，插入建树（BST），如果建完之后的树，它的前序遍历与所给序列相同，那么所给的序列显然是可以构成BST的，应该给出YES，当然所给的序列还应和镜像的遍历比较，如果相同，也可以给出YES。

所以可以用三个数组存（输入序列，树的前序遍历，镜像树的前序遍历），但因为数组不太好比较，用vector更加方便，因为它重载了==符号。

#include<bits/stdc++.h>
using namespace std;
int N;
struct node{
    int data;
    node* lchild;
    node* rchild;
};
void insert(node* &root,int data){
    if(root==NULL){
        root=new node;
        root->data=data;
        root->lchild=root->rchild=NULL;
        return;
    }
    if(data<root->data)insert(root->lchild,data);
    else insert(root->rchild,data);
}
void preOrder(node* root,vector<int>& vi){
    if(root==NULL)return;
    vi.push_back(root->data);
    preOrder(root->lchild,vi);
    preOrder(root->rchild,vi);
}
void preOrderMirror(node* root,vector<int>& vi){
    if(root==NULL)return;
    vi.push_back(root->data);
    preOrderMirror(root->rchild,vi);
    preOrderMirror(root->lchild,vi);
}
void postOrder(node* root,vector<int>& vi){
    if(root==NULL) return;
    postOrder(root->lchild,vi);
    postOrder(root->rchild,vi);
    vi.push_back(root->data);
}
void postOrderMirror(node* root,vector<int>& vi){
    if(root==NULL) return;
    postOrderMirror(root->rchild,vi);
    postOrderMirror(root->lchild,vi);
    vi.push_back(root->data);
}
vector<int> origin,pre,preM,post,postM;
int main(){
    cin>>N;
    int data;
    node* root=NULL;// final root
    for(int i=0;i<N;i++){
        scanf("%d",&data);
        origin.push_back(data);
        insert(root,data);
    }
    preOrder(root,pre);
    preOrderMirror(root,preM);
    postOrder(root,post);
    postOrderMirror(root,postM);
    if(origin==pre){
        printf("YES\n");
        for(int i=0;i<post.size();i++){
            printf("%d",post[i]);
            if(i<post.size()-1)printf(" ");
        }
    }
    else if(origin==preM){
        printf("YES\n");
        for(int i=0;i<postM.size();i++){
            printf("%d",postM[i]);
            if(i<postM.size()-1)printf(" ");
        }
    }
    else{
        printf("NO\n");
    }
    return 0;
}