PAT甲级练习1064. Complete Binary Search Tree (30)

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

完全搜索二叉树的中序遍历就是一个序列的递增排序，反推可以用dfs把值放到对应的位置上，层次遍历用bfs，参考http://blog.csdn.net/jtjy568805874/article/details/50917474

#include <iostream>  #include <cstdio>  #include <algorithm>  #include <vector>  #include <map>  #include <set>  #include <stack>  #include <queue>  #include <string>  #include <string.h> using namespace std;const int MAX = 1e3+10;int n, a[MAX], b[MAX], cnt;void dfs(int x){if(x>n) return;dfs(x<<1);b[x] = a[cnt++];dfs(x<<1|1);}void bfs(int x){queue<int> p; p.push(x);while(!p.empty()){int q = p.front();p.pop();printf("%d%s", b[q], q == n ? "\n" : " ");if((q<<1)<=n) p.push(q<<1);if((q<<1|1)<=n) p.push(q<<1|1);}}int main() {scanf("%d", &n);for(int i=0; i<n; i++){scanf("%d", &a[i]);}sort(a, a+n);dfs(1);bfs(1);scanf("%d",&n);    return 0;}