1043. Is It a Binary Search Tree (25)

题目链接：http://www.patest.cn/contests/pat-a-practise/1043

题目：

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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:

78 6 5 7 10 8 11

Sample Output 1:

YES5 7 6 8 11 10 8

Sample Input 2:

78 10 11 8 6 7 5

Sample Output 2:

YES11 8 10 7 5 6 8

Sample Input 3:

78 6 8 5 10 9 11

Sample Output 3:

NO

分析：

知道前序，其实还知道中序（就是排序后的），然后再看后序。还要注意树中有两个相同的结点的情况，后者把树调换后需要倒着来找rootIdx。

AC代码：

#include<iostream>#include<algorithm>#include<stdio.h>#include<string>using namespace std;struct Node{ int i; Node *lchild; Node *rchild;}Tree[1002];int loc;int str1[1002];//前序int str2[1002];//中序int ans[1002];//ansbool cmp(int x, int y){ return x > y;}Node *creat(){ Tree[loc].lchild = Tree[loc].rchild = NULL; return &Tree[loc++];}int idx = 0;void postOrder(Node *T){ if (T->lchild != NULL){  postOrder(T->lchild); } if (T->rchild != NULL){  postOrder(T->rchild); } ans[idx++] = T->i;}Node* build(int s1, int e1, int s2, int e2){ if (e1 - s1 != e2 - s2)return NULL; Node *ret = creat(); ret->i = str1[s1]; int rootIdx = -1; for (int i = s2; i <= e2; i++){  if (str2[i] == str1[s1]){   rootIdx = i;   break;  } } if (rootIdx == -1)return NULL; if (rootIdx != s2){  ret->lchild = build(s1 + 1, s1 + (rootIdx - s2), s2, rootIdx - 1); } if (rootIdx != e2){  ret->rchild = build(s1 + (rootIdx - s2) + 1, e1, rootIdx + 1, e2); } return ret;}Node* newbuild(int s1, int e1, int s2, int e2){ if (e1 - s1 != e2 - s2)return NULL; Node *ret = creat(); ret->i = str1[s1]; int rootIdx = -1; for (int i = e2; i >= s2; i--){  if (str2[i] == str1[s1]){   rootIdx = i;   break;  } } if (rootIdx == -1)return NULL; if (rootIdx != s2){  ret->lchild = build(s1 + 1, s1 + (rootIdx - s2), s2, rootIdx - 1); } if (rootIdx != e2){  ret->rchild = build(s1 + (rootIdx - s2) + 1, e1, rootIdx + 1, e2); } return ret;}void change(Node *T){ Node *tmp = T->rchild; T->rchild = T->lchild; T->lchild = tmp; if (T->lchild != NULL)change(T->lchild); if (T->rchild != NULL)change(T->rchild);}int main(){ //freopen("F://Temp/input.txt", "r", stdin); int n; cin >> n; for(int i = 0; i < n; i++){  cin >> str1[i];  str2[i] = str1[i]; } sort(str2, str2 + n); loc = 0; Node *T = build(0, n - 1, 0, n - 1); postOrder(T); if (idx < n){//再倒过来找一遍，为了应对有连个相同结点的情况。  sort(str2, str2 + n, cmp );  idx = 0;  T = newbuild(0, n - 1, 0, n - 1);  postOrder(T); } if (idx == n){  cout << "YES" << endl;  for (int i = 0; i < n; i++){   if (i == 0)cout << ans[i];   else cout << " " << ans[i];  } } else cout << "NO"; cout << endl; return 0;}

截图：

——Apie陈小旭