1064. Complete Binary Search Tree (30)

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.

10
1 2 3 4 5 6 7 8 9 0

6 3 8 1 5 7 9 0 2 4

#include <iostream>
#include <vector>
#include<algorithm>
#include<math.h>
#include <queue>
using namespace std;
vector<int> num;
struct Node {
	int val;
	Node *left;
	Node *right;
};
Node* BuildFullTree(int min, int n) {
	if (n == 0) {
		return NULL;
	}
	if (n == 1) {
		Node *p = (Node*)malloc(sizeof(Node));
		p->val = num[min];
		p->left = NULL;
		p->right = NULL;
		return p;
	}
	int level = 1;
	while (true)
	{
		if (pow(2, level) > n) {
			level--;
			break;
		}
		level++;
	}
	int leftCnt, rightCnt;
	if (n + 1 - pow(2, level) < pow(2, level - 1))
		leftCnt = n - pow(2, level - 1);
	else
		leftCnt = pow(2, level) - 1;
	rightCnt = n - 1 - leftCnt;
	Node *p = (Node*)malloc(sizeof(Node));
	p->val = num[min+leftCnt];
	p->left = BuildFullTree(min, leftCnt);
	p->right = BuildFullTree(min+leftCnt + 1, rightCnt);
	return p;
}
int main(void) {
	int n;
	cin >> n;
	for (int i = 0; i < n; i++) {
		int temp;
		cin >> temp;
		num.push_back(temp);
	}
	sort(num.begin(), num.end());
	Node *root = (Node*)malloc(sizeof(Node));
	root = BuildFullTree(0, n);
	queue<Node*> q;
	q.push(root);
	while (true)
	{
		if (q.front()->left != NULL)
			q.push(q.front()->left);
		if (q.front()->right != NULL)
			q.push(q.front()->right);
		cout << q.front()->val;
		q.pop();
		if (q.empty())
			break;
		cout << " ";
	}
	return 0;
}