The order of a Tree
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1749 Accepted Submission(s):
908
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
给一个序列,依照序列建二叉排序树,输出建成这棵树的最小字典序列。
不能使用线段树那种表示方式来表示树,因为可能会退化为一条链,这时会溢出。
所以用lson和rson两个数组分别存每个结点的左右儿子(题中规定,数字为1—n)。
#include<iostream> #include<cstdio> #include<cstring> #include<queue> using namespace std; int lson[100005],rson[100005],n; void Insert(int rt,int x) { if(rt>x) { if(lson[rt]==0) lson[rt]=x; else Insert(lson[rt],x); } else { if(rson[rt]==0) rson[rt]=x; else Insert(rson[rt],x); } } int cnt=0; void order(int rt) //深搜 { if(rt==0) return; printf("%d",rt); cnt++; if(cnt==n) printf(" "); else printf(" "); order(lson[rt]); order(rson[rt]); } int main() { while(scanf("%d",&n)!=EOF) { cnt=0; int num,root; memset(lson,0,sizeof(lson)); memset(rson,0,sizeof(rson)); scanf("%d",&root); for(int i=1; i<n; i++) { scanf("%d",&num); Insert(root,num); } order(root); //cout<<n<<" "<<cnt<<endl; } return 0; }