1064. Complete Binary Search Tree (30)

最新推荐文章于 2019-09-21 22:10:36 发布

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本文介绍如何通过给定的一组非负整数构建一个既为二叉搜索树又为完全二叉树的数据结构，并输出该树的层级遍历序列。文章详细说明了输入输出规范及实现步骤。

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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:

10
1 2 3 4 5 6 7 8 9 0

Sample Output:

6 3 8 1 5 7 9 0 2 4

题目的整体思路就是建立有n个节点的完全二叉树（使用广度优先搜索的思路），将给出的n个数值按升序排列，然后用中序遍历将这些数值分配到对应的完全二叉树的节点上，最后再广度优先搜索分层输出就行（只要按照节点的序号进行输出就行）。

#include <iostream>
#include <vector>
#include <algorithm>
#include <map>
#include <queue>
using namespace std;
vector<int> keys;
map<int,vector<int> >tree;
void inor_traversal(int root);
int num,cnt=0;
int main()
{
    cin >>num;
    for(int i=0;i<num;i++)
    {
        int tmp;
        cin >>tmp;
        keys.push_back(tmp);
    }
    sort(keys.begin(),keys.end());
    queue<int> q;
    int tmp,i=0;
    q.push(i);
    while(!q.empty())
    {
        tmp=q.front();
        q.pop();
        i++;
        if(i<num)
        {
            tree[tmp].push_back(i);
            q.push(i);
        }
        else
            tree[tmp].push_back(-1);
        i++;
        if(i<num)
        {
            tree[tmp].push_back(i);
            q.push(i);
        }
        else
            tree[tmp].push_back(-1);
    }
    inor_traversal(0);
    i=0;
    cout <<tree[i].at(2);
    for(int i=1;i<num;i++)
        cout <<" "<<tree[i].at(2);
    cout <<endl;
    return 0;
}
void inor_traversal(int root)
{
    int tmp=tree[root].at(0);
    if(tmp!=-1)
        inor_traversal(tmp);
    tree[root].push_back(keys[cnt]);
    cnt++;
    tmp=tree[root].at(1);
    if(tmp!=-1)
        inor_traversal(tmp);
    return;
}