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

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 $N(\le 1000)$. 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

---

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <stdlib.h>
#define MAXN 1000

//按顺序输出结果（层次遍历）
void Output(int T[], int n) {
	for (int i = 0; i < n; i++) {
		if (i == 0) printf("%d", T[i]);
		else printf(" %d", T[i]);
	}
	printf("\n");
}

//获取一个含有n个结点的完全二叉树的层数
int GetLevel(int n) {
	int count = 0;
	while (n) {
		count++;
		n /= 2;
	}
	return count;
}

//获取一个n层的满二叉树的总结点数
int GetSum(int level) {
	if (level < 1) return 0;
	int sum = 1;
	for (int i = 0; i < level; i++) sum *= 2;
	return --sum;
}

//获取一个满二叉树第n层结点的个数
int GetLevelSum(int level) {
	if (level < 1) return 0;
	int sum = 1;
	for (int i = 0; i < level - 1; i++) sum *= 2;
	return sum;
}

//获取一个含有n个结点的完全二叉树的左子树的节点数
int GetLeftLength(int n) {
	int Level = GetLevel(n), sum = 0;
	sum += GetSum(Level - 2);
	sum += (n - GetSum(Level - 1) < GetLevelSum(Level) / 2) ? n - GetSum(Level - 1) : GetLevelSum(Level) / 2;
	return sum;
}

//递归求解
void solve(int input[], int tree[], int inputLeft, int inputRight, int treeRoot) {
	int n = inputRight - inputLeft + 1, L, leftTreeRoot, rightTreeRoot;
	if (n == 0) return;
	L = GetLeftLength(n);
	tree[treeRoot] = input[inputLeft + L];
	leftTreeRoot = treeRoot * 2 + 1;
	rightTreeRoot = leftTreeRoot + 1;
	solve(input, tree, inputLeft, inputLeft + L - 1, leftTreeRoot);
	solve(input, tree, inputLeft + L + 1, inputRight, rightTreeRoot);
}

//比较函数，qsort函数会用到
int compare(const void* a, const void* b) {
	return *(int*)a - *(int*)b;
}

int main(void) {
	int n, input[MAXN], tree[MAXN];
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		scanf("%d", &input[i]);
	}
	//对输入序列按从大到小的顺序排序
	qsort(input, n, sizeof(int), compare);
	//递归求解
	solve(input, tree, 0, n - 1, 0);
	Output(tree, n);
	return 0;
}