04-树6 Complete Binary Search Tree (30 分)

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 (≤). 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

 

一、题目要求将输入的值用中序遍历然后在用层序遍历输出，所以中序遍历只需要先排列，然后找到最中间的值就行。加上完全树的特性，如果根为0,节点i，左边的儿子是2*i+1，右边的儿子2*i+2。将其排序好后，用数组和层序遍历将其储存和排列最后输出。

二、代码

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#define MAX 1005
int N, n[MAX], ins[MAX], k = 0;
int cmp(const int *a, const int *b)
{
    return *a - *b;
}
void inorder(int root);
int main()
{
    int i;
    scanf("%d", &N);
    for (i = 0; i < N; i++) scanf("%d", &n[i]);
    qsort(n, N, sizeof(int), cmp);
    inorder(0);
    printf("%d", ins[0]);
    for (i = 1; i < N; i++) printf(" %d", ins[i]);
}
void inorder(int root)
{
    if (root >= N) return;
    inorder(2 * root+1);
    ins[root] = n[k++];
    inorder(2 * root + 2);
}

本文参考了：https://blog.csdn.net/qq_36888550/article/details/87929937

　　　　　　https://blog.csdn.net/ryo_218/article/details/81462130

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