1064. Complete Binary Search Tree (30)

Posted 2020-10-29 Dyleaf

1064. Complete Binary Search Tree (30)

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

#include <bits/stdc++.h>
using namespace std;

const int maxn = 1000;
int node[maxn],level[maxn];

void levelorder(int left, int right, int idx){
    if(left>right) return;
    int n = right -left+1;
    int l = log(n+1)/log(2);
    int leave = n-pow(2,l)+1;
    int root = left + pow(2, l-1)-1 +min((int)pow(2, l-1), leave);
    // througu idx to get leave order;
    level[idx] = node[root];
    levelorder(left, root-1, idx*2);
    levelorder(root+1, right, idx*2+1);
}
int main(){
    int n;
    scanf("%d", &n);
    for(int i=1; i<= n; i++){
        scanf("%d", &node[i]);
    }
    sort(node+1, node+n+1);
    levelorder(1,n, 1);
    for(int i=1; i<=n ;i++){
        if(i!=1) printf(" ");
        printf("%d",level[i]);
    }
    return 0;
}