HDU 3999 The order of a Tree (先序遍历)
遍历 The of HDU Tree order 先序
2023-09-27 14:22:23 时间
The order of a Tree
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 835 Accepted Submission(s): 453
Problem Description
As we know,the shape of a binary search tree is greatly related to the order of keys we insert. To be precisely:
1. insert a key k to a empty tree, then the tree become a tree with
only one node;
2. insert a key k to a nonempty tree, if k is less than the root ,insert
it to the left sub-tree;else insert k to the right sub-tree.
We call the order of keys we insert “the order of a tree”,your task is,given a oder of a tree, find the order of a tree with the least lexicographic order that generate the same tree.Two trees are the same if and only if they have the same shape.
1. insert a key k to a empty tree, then the tree become a tree with
only one node;
2. insert a key k to a nonempty tree, if k is less than the root ,insert
it to the left sub-tree;else insert k to the right sub-tree.
We call the order of keys we insert “the order of a tree”,your task is,given a oder of a tree, find the order of a tree with the least lexicographic order that generate the same tree.Two trees are the same if and only if they have the same shape.
Input
There are multiple test cases in an input file. The first line of each testcase is an integer n(n <= 100,000),represent the number of nodes.The second line has n intergers,k1 to kn,represent the order of a tree.To make if more simple, k1 to kn is a sequence of 1 to n.
Output
One line with n intergers, which are the order of a tree that generate the same tree with the least lexicographic.
Sample Input
4
1 3 4 2
Sample Output
1 3 2 4
Source
Recommend
lcy
题意是给你一个序列建立一棵二叉搜索树 要你找出另外一个序列可以建立和原序列建立的二叉搜索树一样且这个序列是字典序最小,一看根据二叉搜索树的特点 左孩子小 右孩子大 直接输出一个先序遍历就OK!
#include<iostream> #include<cstdio> #include<cstring> #include<stack> #include<cstdlib> using namespace std; const int N=1010; struct Tree{ Tree *l,*r; int x; }tree; Tree *root; Tree *Create(Tree *rt,int x){ if(rt==NULL){ rt=(Tree *)malloc(sizeof(Tree)); rt->x=x; rt->l=rt->r=NULL; return rt; } if(rt->x>x) //insert a key k to a nonempty tree, if k is less than the root ,insert it to the left sub-tree rt->l=Create(rt->l,x); else //else insert k to the right sub-tree rt->r=Create(rt->r,x); return rt; } void PreOrder(Tree *rt,int x){ //先序历遍 if(x==1) printf("%d",rt->x); else printf(" %d",rt->x); if(rt->l!=NULL) PreOrder(rt->l,2); if(rt->r!=NULL) PreOrder(rt->r,2); } int main(){ //freopen("input.txt","r",stdin); int n,x; while(~scanf("%d",&n)){ root=NULL; for(int i=0;i<n;i++){ scanf("%d",&x); root=Create(root,x); } PreOrder(root,1); printf("\n"); } return 0; }
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