[数据结构学习日记]PTA-Complete Binary Search Tree

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

#include 
#include 
#include 
#include 
using std::cin;
using std::cout;
using std::vector;
using std::min;
void solve(vector a, int Left, int Right, int* b, int root) {
	int N = Right - Left + 1;
	if (N <= 0) {
		return;
	}
	int level = static_cast  (log2(N+1));//计算完全二叉树的层数；不计未满的层数；
	int nums = N - (pow(2, level) - 1);//用于计算未满层数结点的个数；
	int L = pow(2, level - 1) - 1 + min(static_cast (pow(2, level - 1)) , nums);
	int R = N - L - 1;
	b[root] = a[Left + L];
	int Lroot = root * 2 + 1;
	int Rroot = Lroot + 1;
	solve(a, Left, Left + L - 1, b, Lroot);
	solve(a, Left + L + 1, Left + L  + R, b, Rroot);
}
int main() {
	int N; //数据个数
	cin >> N;
	vector a; //用于存放原始数据
	int* b = new int[N];//用于存放树状结构；
	for (int i = 0; i < N; i++) {
		int num;
		cin >> num;
		a.push_back(num);
	}
	sort(a.begin(), a.end());
	solve(a, 0, N - 1, b, 0);
	int ret = 1;
	for (int i = 0; i < N; i++) {
		if (ret == 1) {
			cout << b[i];
			ret = 0;
		}
		else {
			cout << ' ' << b[i];
		}
	}
	return 0;
}