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
solution:
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef struct node *tree;
struct node
{
int val;
tree left,right;
};
int cnt=0;
void built(tree root,vector<int>&num)
{
if(root==NULL)return;
built(root->left,num);
root->val=num[cnt++];
built(root->right,num);
}
int main()
{
int n;cin>>n;
tree root=new node;
root->left=root->right=NULL;
root->val=1;
function<void(tree)> build=[&](tree root)->void
{
if(root==NULL)return;
if(root->val*2<=n)
{
tree l=new node;
l->left=l->right=NULL;
l->val=root->val*2;
root->left=l;
}
if(root->val*2+1<=n)
{
tree r=new node;
r->left=r->right=NULL;
r->val=root->val*2+1;
root->right=r;
}
build(root->left),build(root->right);
};
build(root);
vector<int>a(n);
for(int i=0;i<n;i++)cin>>a[i];
sort(a.begin(),a.end());
built(root,a);
queue<tree>q;
q.push(root);
vector<int>ans;
while(!q.empty())
{
tree tmp=q.front();
q.pop();
ans.push_back(tmp->val);
if(tmp->left)q.push(tmp->left);
if(tmp->right)q.push(tmp->right);
}
for(int i=0;i<ans.size();i++)
{
if(i)cout<<' ';
cout<<ans[i];
}
}
大致思路是,先建立完全二叉树的框架,其中的数字是什么无所谓,之后将题目给出的数据经过排序后按中序遍历的方式插入完全二叉树。