【PAT】1064. Complete Binary Search Tree (30)

题目链接：http://pat.zju.edu.cn/contests/pat-a-practise/1064

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：

（1）难点在这个数组中确定根节点。

（2）用queue进行层遍历。

参考代码：

#include<iostream>
#include<algorithm>
#include<queue>
#include<cmath>
using namespace std;

struct TreeNode
{
	int value;
	TreeNode * left;
	TreeNode * right;
	TreeNode(int v):value(v),left(NULL),right(NULL){}
	TreeNode():left(NULL),right(NULL){}
};

void build(int *array,int len, TreeNode **r)
{
	int level = 0;
	int i;
	for(i=0; i<100; i++)
	{
		//n层满二叉树的节点个数是  pos(2.0,i);
		if((int)pow(2.0,i) -1 > len)
		{
			level = i;  break;
		}
	}
	int remain; //最后一层节点的个数
	int left; //左子树的节点个数
	int right; //右子树的节点个数
	int last; //最后一层最左边节点的编号
	int c;//满二叉树中节点的个数
	c = (int)pow(2.0,level-1) - 1;
	remain = len - c ;
	last = c + 1;
	// last/2 代表根节点的左子树在最后一层上可容纳的节点个数
	if(remain > last/2){
		left = (c-1)/2 + last/2;		
	}else{
		left = (c-1)/2 + remain;
	}
	right = len - left - 1;
	
	TreeNode *root = new TreeNode();
	*r = root;
	root->value = array[left];
	if(left > 0)
		build(array, left, &root->left);
	if(right > 0)
		build(array+left+1, right, &root->right);
}

int main()
{
	int n,i;
	while(cin>>n)
	{
		int *array = new int[n];		
		for(i=0; i<n; i++)
			cin>>array[i];
		sort(array,array+n);
		TreeNode *root = NULL;
		build(array,n,&root);
		queue<TreeNode *> que;		
		que.push(root);
		bool flag = false;
		TreeNode *temp;
		while(!que.empty()){
			if(flag)
				cout<<" ";
			else
				flag = true;
			temp = que.front();
			if(temp){
				cout<<temp->value;
				if(temp->left)
					que.push(temp->left);
				if(temp->right)
					que.push(temp->right);
			}
			que.pop();
		}
		cout<<endl;
	}
	return 0;
}