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 NN (\le 1000≤1000). Then NN 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
思路: 这个题想了好久,最后还是看别人的代码才明白了,就是根据完全二叉树的性质,完全二叉树在数组中某个节点的下标为i,则其左右孩子的下标分别为2*1,2*i+1。然后递归建树。
AC代码:
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
int pos = 0;
int root = 1;
int N;
vector<int> CBT;
vector<int> AVL;
void Build_CBT(int root)
{
if ( root > N){
return;
}else{
int left = 2 * root;
int right = 2 * root + 1;
Build_CBT(left);
CBT[root] = AVL[pos++];
Build_CBT(right);
}
}
int main()
{
cin>>N;
AVL = vector<int>(N,0);
CBT = vector<int>(N+1,0);
for ( int i = 0 ; i < N ; i++){
cin>>AVL[i];
}
sort(AVL.begin(),AVL.end());
Build_CBT(root);
cout<<CBT[1];
for ( int i = 2 ; i < CBT.size() ; i++){
cout<<" "<<CBT[i];
}
return 0;
}