题目如下：
A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties:
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 11Sample Output 1:
YES 5 7 6 8 11 10 8Sample Input 2:
7 8 10 11 8 6 7 5Sample Output 2:
YES 11 8 10 7 5 6 8Sample Input 3:
7 8 6 8 5 10 9 11Sample Output 3:
NO
第一步是建立BST，建立方法很简单，首先定义一个Node为空来存储根结点，定义一个insert方法，当T为空时建立该结点，当T不空时，根据BST条件对T->left和T->right进行递归。
void insert(Node &T,int num){
if(!T){
T = new TreeNode();
T->data = num;
return;
}
if(num < T->data){
insert(T->left,num);
}else{
insert(T->right,num);
}
}需要注意的是Node参数应当传入引用，因此加一个&，否则无法直接操作到外部的T参数。
对于整个输入序列，调用N次insert方法即可得到完整的BST，且根结点为T。
下面只需要对BST进行一次前序遍历和一次镜像前序遍历（根右左），看题目给出的序列和哪个相符，可以判断出是BST还是MBST，如果都不是则输出NO。
完整代码如下：
#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <vector>
using namespace std;
typedef struct TreeNode* Node;
struct TreeNode{
Node left;
Node right;
int data;
TreeNode(){
left = right = NULL;
}
};
int *pre;
int N;
vector<int> preOrderV;
vector<int> mpreOrderV;
vector<int> result;
void insert(Node &T,int num){
if(!T){
T = new TreeNode();
T->data = num;
return;
}
if(num < T->data){
insert(T->left,num);
}else{
insert(T->right,num);
}
}
void preOrder(Node T){
if(!T) return;
preOrderV.push_back(T->data);
preOrder(T->left);
preOrder(T->right);
}
void postOrder(Node T){
if(!T) return;
postOrder(T->left);
postOrder(T->right);
result.push_back(T->data);
}
void MBSTpreOrder(Node T){
if(!T) return;
mpreOrderV.push_back(T->data);
MBSTpreOrder(T->right);
MBSTpreOrder(T->left);
}
bool arrayEqual2Vector(int *arr, vector<int> vec){
for(int i = 0; i < vec.size(); i++){
if(arr[i] != vec[i]) return false;
}
return true;
}
void mirror(Node T){
if(!T)return;
Node temp = T->left;
T->left = T->right;
T->right = temp;
mirror(T->left);
mirror(T->right);
}
int main()
{
cin >> N;
pre = (int*)malloc(sizeof(int)*N);
int num;
Node T = NULL;
for(int i = 0; i < N; i++){
scanf("%d",&num);
pre[i] = num;
insert(T,num);
}
preOrderV.clear();
mpreOrderV.clear();
result.clear();
preOrder(T);
MBSTpreOrder(T);
if(arrayEqual2Vector(pre,preOrderV)){
cout << "YES" << endl;
postOrder(T);
}else if(arrayEqual2Vector(pre,mpreOrderV)){
cout << "YES" << endl;
mirror(T);
postOrder(T);
}else{
cout << "NO" << endl;
}
if(result.size() != 0){
cout << result[0];
for(int i = 1; i < result.size(); i++)
printf(" %d",result[i]);
cout << endl;
}
return 0;
}
版权声明：本文为博主原创文章，未经博主允许不得转载。
1043. Is It a Binary Search Tree (25)
原文地址：http://blog.csdn.net/xyt8023y/article/details/46849933
