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 <stdio.h>
void sort(int *keys, int n) {			//元素并不多，直接用插入排序
	for (int i = 1; i < n; ++i) {
		int temp = keys[i];
		int j = i - 1;
		while (j >= 0 && keys[j] > temp) {
			keys[j + 1] = keys[j];
			--j;
		}
		keys[j + 1] = temp;
	}
}
int main() {
	freopen("test.txt", "r", stdin);
	int n;
	scanf("%d", &n);
	int keys[1000] = {};
	for (int i = 0; i < n; ++i)
		scanf("%d", &keys[i]);
	sort(keys, n);						//对关键字进行排序
	int CBT[1001] = {};					//层序顺序保存树，1位置为树根，则i的左右儿子分别为2*i和2*i+1; 类似堆
	int heap[1001] = {};				//用于中序遍历的堆
	int root = 1, k = 0, heapSize = 0;	//k：已遍历元素的个数；  heapSize：堆中元素个数
	//中序遍历的非递归实现过程；因为中序遍历搜索时可以得到各节点的排序，所以根据中序遍历顺序依次填充关键字可以得到答案
	while (root <= n || heapSize) {		//树或者堆部位空
		while (root <= n) {				//沿着左子树路径找到最左叶节点，并将路径上的所有节点压栈
			heap[heapSize++] = root;	//入栈
			root *= 2;					//指向左子树
		}
		if (heapSize) {					
			root = heap[--heapSize];	//弹出栈顶元素
			CBT[root] = keys[k++];		//填充当前最小关键字
			root = 2 * root + 1;		//转向右子树
		}
	}
	for (int i = 1; i <= n; ++i) {		//按照层序顺序输出CBT
		if (i != 1)
			printf(" ");
		printf("%d", CBT[i]);
	}

	return 0;
}

题目链接：http://www.patest.cn/contests/mooc-ds/04-%E6%A0%918

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