The order of a Tree

最新推荐文章于 2021-11-16 00:30:26 发布

原创最新推荐文章于 2021-11-16 00:30:26 发布 · 203 阅读

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博客围绕二叉搜索树展开，指出树的形状与插入键的顺序密切相关。给定一个生成二叉树的键顺序，任务是找出能生成相同形状树且字典序最小的键顺序。解题思路是先构造二叉搜索树，再进行先序遍历。

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

题意：给出一个n，接下来n个数是按一定的序列构成了一棵二叉树，求一个序列既能和这个二叉树相同又字典序最小。

思路：首先构造出这个二叉搜索树，然后先序便利即可。代码如下:

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define N 100010
using namespace std;
int a[N],le[N],r[N],sum[N];
int l1,l2,num;
void build(int l,int k)
{
    if(a[l]>=k)
    {
        if(le[l])
            build(le[l],k);
        else
            le[l]=l2;
    }
    else
    {
        if(r[l])
            build(r[l],k);
        else
            r[l]=l2;
    }
}
void check(int l)
{
    sum[num++]=a[l];
    if(le[l])
        check(le[l]);
    if(r[l])
        check(r[l]);
}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        int i,j,k;
        memset(le,0,sizeof(le));
        memset(r,0,sizeof(r));
        l1=-1,l2=0,num=0;
        for(i=0; i<n; i++)
        {
            scanf("%d",&k);
            if(l1==-1)
                a[++l1]=k;
            else
            {
                a[++l2]=k;
                build(l1,k);
            }
        }
        check(l1);
        for(i=0; i<num-1; i++)
            printf("%d ",sum[i]);
        printf("%d\n",sum[num-1]);
    }
    return 0;
}