PAT (Advanced Level) Practise 1064 Complete Binary Search Tree (30)

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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

题意:给你n个数,让你将他们插入一棵完全二叉搜索树中

解题思路:很明显将所有数从小到大排序后,这个序列就是这棵树的中序遍历,dfs插入即可

#include <iostream> #include <cstdio> #include <cstring> #include <string> #include <algorithm> #include <queue> #include <stack> #include <cmath> #include <map> #include <bitset> #include <set> #include <vector> #include <functional> using namespace std; #define LL long long const int INF = 0x3f3f3f3f; int n,a[1005],cnt,b[1005]; void dfs(int k) { if(k>n) return ; dfs(k*2); b[k]=a[cnt++]; dfs(k*2+1); } int main() { while(~scanf("%d",&n)) { for(int i=1;i<=n;i++) scanf("%d",&a[i]); sort(a+1,a+1+n); cnt=1; dfs(1); printf("%d",b[1]); for(int i=2;i<=n;i++) printf(" %d",b[i]); printf("\n"); } return 0; }

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