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  • HDU 3999 The order of a Tree (先序遍历)

    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.
     
    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("
    ");
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/jackge/p/3222903.html
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