1064. Complete Binary Search Tree (30)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties:

• The left subtree of a node contains only nodes with keys less than the node's key.
• The right subtree of a node contains only nodes with keys greater than or equal to the node's key.
• Both the left and right subtrees must also be binary search trees.

A Complete Binary Tree (CBT) is a tree that is completely filled, with the possible exception of the bottom level, which is filled from left to right.

Now given a sequence of distinct non-negative integer keys, a unique BST can be constructed if it is required that the tree must also be a CBT. You are supposed to output the level order traversal sequence of this BST.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (<=1000). Then N distinct non-negative integer keys are given in the next line. All the numbers in a line are separated by a space and are no greater than 2000.

Output Specification:

For each test case, print in one line the level order traversal sequence of the corresponding complete binary search tree. All the numbers in a line must be separated by a space, and there must be no extra space at the end of the line.

Sample Input:

101 2 3 4 5 6 7 8 9 0

Sample Output:

6 3 8 1 5 7 9 0 2 4

思路：该题是给出N个数的序列，然后将序列构成一个完全搜索二叉树

由于树的中序遍历是从小到大输出的，因此可以模拟中序遍历将序列从小到大排列好后一个一个填到树中去

因为是完全二叉搜索数，所以可以用数组下标直接表示树的根结点，若用0作为根结点，那每个根结点的左右子树分别为root * 2 + 1和root * 2 + 2

#include<stdio.h>#include<stdlib.h> #include<math.h>#include<string.h>#include<algorithm>#include <vector> #include<stack> #include <queue>  #include<map>#include<iostream>  #include <functional> #define MAX 110#define MAXD 10001#define TELNUM 10using namespace std;int num[2001];int num1[2001];int n,k = 0;void tian(int root){  if(root > n - 1)  {    return;  }  tian(root * 2 + 1);  num1[root] = num[k++];  tian(root * 2 + 2);  }int main(void){  scanf("%d",&n);    for(int i = 0; i < n; i++)  {    scanf("%d",&num[i]);  }    sort(num,num + n);    tian(0);    for(int i = 0; i < n; i++)  {    printf("%d",num1[i]);    if(i < n - 1)    {      printf(" ");    }  }    return 0;}