Pat(Advanced Level)Practice--1064(Complete Binary Search Tree)

Pat1064代码

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:

101 2 3 4 5 6 7 8 9 0

Sample Output:

6 3 8 1 5 7 9 0 2 4

AC代码：

本题同时满足完全二叉树和二叉搜索数的性质，因此不必去重建二叉树；因为我们将节点的值从小到大排列之后，依次是left子树->root->right子树节点的值。

#include<cstdio>#include<algorithm>#define MAX 1005using namespace std;int Keys[MAX];int Level[MAX];int ID=0;void ReBulid(int i,int n){int l=2*i;int r=2*i+1;if(i>n)return;ReBulid(l,n);Level[i]=Keys[ID++];if(r<=n)ReBulid(r,n);}int main(int argc,char *argv[]){int i,n;scanf("%d",&n);for(i=0;i<n;i++)scanf("%d",&Keys[i]);sort(Keys,Keys+n);ReBulid(1,n);for(i=1;i<n;i++)printf("%d ",Level[i]);printf("%d\n",Level[n]);return 0;}