1064. Complete Binary Search Tree (30)二叉树——PAT (Advanced Level) Practise

Posted 2020-09-29 liguangsunls

题目信息

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

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

解题思路

模拟建树推出层序

AC代码

#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
int n, a[1005];
vector<int> level[22];
void step(int loc, int len, int lv){
  if (len <= 0) return;
  int t = 1;
  while (t*2 <= len) t *= 2;
  --t;
  int cd = ((len - (len - t)) + 1) / 2 + len - t;

  if (cd <= t + 1){
    level[lv].push_back(a[loc + cd - 1]);
    step(loc, cd - 1, lv + 1);
    step(loc + cd, len - cd, lv + 1);
  }else{
    level[lv].push_back(a[loc + t]);
    step(loc, t, lv + 1);
    step(loc + t + 1, len - t - 1, lv + 1);
  }
}
int main()
{
  scanf("%d", &n);
  for (int i = 0; i < n; ++i){
    scanf("%d", a+i);
  }
  sort(a, a + n);
  step(0, n, 0);
  printf("%d", level[0][0]);
  for (int i = 1; i <= 13; ++i){
    for (int j = 0; j < level[i].size(); ++j){
      printf(" %d", level[i][j]);
    }
  }
  printf("\n");
}

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