04-树8. Complete Binary Search Tree (30)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

8000 B

判题程序

Standard

作者

CHEN, Yue

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:

101 2 3 4 5 6 7 8 9 0

Sample Output:

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CBTif (i != 1)printf(" ");printf("%d", CBT[i]);}return 0;}

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