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.
拜托，一个“&”没写硬是折腾了一下午，指针神马的真的是生人勿进，有测试点通不过就再说了，exhuasted……
#include <stdio.h>#include <stdlib.h>#include <vector>using namespace std;#define N 1001typedef struct Node{int id;struct Node *left;struct Node *right;}Node;vector<int> num;//存放初始序列vector<int> pre;//存放前序序列 vector<int> prem;//存放前序镜像序列 vector<int> post;vector<int> postm;int n;void Postm (Node * root,vector<int> & vi);void Post (Node * root,vector<int> & vi);void Pre (Node * root,vector<int>& vi);void Prem (Node * root,vector<int>& vi);void Insert (Node * &root,int n);int main (){int temp,i;scanf("%d",&n);Node * root=NULL;for( i=0;i<n;i++){scanf("%d",&temp);num.push_back(temp);Insert(root,temp);}//构造完成//先看是否是前序序列Pre(root,pre);if( num==pre)//是前序序列{printf("YES\n");Post(root,post);printf("%d",post[0]);for( i=1;i<post.size();i++) printf(" %d",post[i]);printf("\n");}else{Prem(root,prem);//是否是前序镜像序列;if( prem==num){printf("YES\n");Postm(root,postm);printf("%d",postm[0]);for( i=1;i<postm.size();i++) printf(" %d",postm[i]);printf("\n");}else printf("No\n");}system("pause");return 0;}void Postm (Node * root,vector<int> & vi){if( root==NULL) return;Postm(root->right,vi);Postm(root->left,vi);vi.push_back(root->id);}void Post (Node * root,vector<int>& vi){if( root==NULL) return;Post(root->left,vi);Post(root->right,vi);vi.push_back(root->id);}void Prem (Node * root,vector<int>& vi){if( root==NULL) return;vi.push_back(root->id);Prem( root->right,vi);Prem( root->left,vi);}void Pre (Node * root,vector<int>& vi){if( root==NULL) return;vi.push_back(root->id);Pre( root->left,vi);Pre( root->right,vi);}void Insert (Node * &root,int n){if( root==NULL){root=new Node;root->id=n;root->left=NULL;root->right=NULL;return;}if( n<root->id) Insert(root->left,n);else Insert(root->right,n);}
，就得加倍付出汗水，赢得场场精彩
