题目描述

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

如果只是构建一个二叉搜索树那么可以一个一个地插入，但现在是构建完全二叉搜索树，这个方法便不行了。联想到二叉搜索树中序遍历是递增的这一个特点，我们可以通过中序遍历来建树。因此我们需要先将输入的序列排序，然后通过排序后的序列来建树。由于题目要求层次遍历输出序列，用顺序结构保存树更简单。

代码

#include <stdio.h>

#define MAXSIZE 1000

//数组a存递增序列，数组b存层次遍历序列
int a[MAXSIZE], b[MAXSIZE];

void quickSort(int low, int high);
void inorder(int root, int n);

int main() {
    int N;
    scanf("%d", &N);
    if (N) {
        for (int i = 0; i < N; i++) {
            scanf("%d", &a[i]);
        }
        quickSort(0, N - 1);//让数组a递增
        inorder(0, N);//中序遍历数组b，并为b赋值
        printf("%d", b[0]);
        for (int i = 1; i < N; i++) {
            printf(" %d", b[i]);
        }
    }
    return 0;
}

//对数组a快速排序
void quickSort(int low, int high) {
    if (low >= high) return;
    int tmp = a[low];
    int i = low, j = high;
    while (i < j) {
        while (i < j && a[j] >= tmp) j--;
        a[i] = a[j];
        while (i < j && a[i] <= tmp) i++;
        a[j] = a[i];
    }
    a[i] = tmp;
    quickSort(low, i - 1);
    quickSort(i + 1, high);
}

//对数组b中序遍历
int index = 0;
void inorder(int root, int n) {
    if (root >= n) return;
    int left = 2 * root + 1;
    int right = left + 1;
    inorder(left, n);
    b[root] = a[index++];//由于a是递增的，因此可以利用a为b赋值
    inorder(right, n);
}