HDU 3999 The order of a Tree

最新推荐文章于 2020-05-29 19:23:42 发布

weixin_30729609

最新推荐文章于 2020-05-29 19:23:42 发布

阅读量67

点赞数

CC 4.0 BY-SA版权

文章标签： java

原文链接：http://www.cnblogs.com/stepping/p/5513236.html

本文探讨了如何根据给定的二叉搜索树构建顺序，找出生成相同形状树的最小字典序构建顺序。通过递归算法实现了这一过程，并提供了完整的代码实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

The order of a Tree

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 66 Accepted Submission(s) : 35

Font: Times New Roman | Verdana | Georgia

Font Size: ← →

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.

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

2011 Multi-University Training Contest 16 - Host by TJU

#include <iostream>
#include<cstdio>
using namespace std;
int ans[100005];
int i,n,k,l;
struct node
{
    int num,left,right;
}tree[100005];

void work(int k)
{
    if (k==-1) return;
    ans[++l]=tree[k].num;
    if (tree[k].left!=-1) work(tree[k].left);
    if (tree[k].right!=-1) work(tree[k].right);
}
int main()
{
    for(i=1;i<=100005;i++)
    {
        tree[i].num=-1;
        tree[i].left=-1;
        tree[i].right=-1;
    }
    scanf("%d",&n);
    if (n>=1)
        {
            scanf("%d",&k); tree[1].num=k;
            for (i=2;i<=n;i++)
            {
                scanf("%d",&k);
                tree[i].num=k;
                int t=1;
                while(1)
                {
                    while (k>tree[t].num && tree[t].right!=-1) t=tree[t].right;
                    if (k>tree[t].num && tree[t].right==-1)
                    {
                        tree[t].right=i;
                        break;
                    }
                    while (k<tree[t].num && tree[t].left!=-1) t=tree[t].left;
                    if (k<tree[t].num && tree[t].left==-1)
                    {
                        tree[t].left=i;
                        break;
                    }
                }
            }
            l=0;
            work(1);
            for(i=1;i<l;i++) printf("%d ",ans[i]);
            printf("%d\n",ans[l]);
        }
    return 0;
}

转载于:https://www.cnblogs.com/stepping/p/5513236.html