题目描述
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.
输入描述:
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.
输出描述:
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.
输入例子:
10
1 2 3 4 5 6 7 8 9 0
输出例子:
6 3 8 1 5 7 9 0 2 4
将一串数字按照从小到大的顺序排列就是中序排序,本题主要考察如何将中序排序转化为层次遍历,先访问根节点,再依次访问左右结点
此模板应熟记于心。
满分代码如下:
#include<bits/stdc++.h>
using namespace std;
const int N=1010;
int a[N],tree[N];
int n;
void inorder(int root,int &id){
if(root>n) return;
inorder(2*root,id);
tree[root]=a[id];
id++;
inorder(2*root+1,id);
}
int main(){
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin>>n;
for(int i=1;i<=n;i++){
cin>>a[i];
}
sort(a+1,a+n+1);
int idd=1,root=1;
inorder(root,idd);
for(int i=1;i<=n;i++){
if(i!=1) cout<<" ";
cout<<tree[i];
}
return 0;
}