1064 Complete Binary Search Tree (30 分)

1064 Complete Binary Search Tree (30 分)  

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 (≤). 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这个二叉树挺好的，我用的是bfs遍历写的。我看到其他的博客都是建树，我比较懒（逃。。。）算是分治的写法了。 1 #include <bits/stdc++.h> 2 3 using namespace std; 4 int n; 5 int an[3000]; 6 struct Node 7 { 8 int left, right; 9 }; 10 int val[11] = {1,2,4,8,16,32,64,128,256,512,1024}; 11 queue<Node> q; 12 vector<int> v; 13 int main(){ 14 cin >> n; 15 for(int i = 0 ; i < n; i++){ 16 cin >> an[i]; 17 } 18 sort(an, an+n); 19 q.push({0,n-1}); 20 while(!q.empty()){ 21 Node node = q.front(); 22 q.pop(); 23 int ll = node.left, rr = node.right; 24 int len = rr-ll+1; 25 int pos = 0; 26 while(val[pos] <= len){ 27 len -= val[pos]; 28 pos ++; 29 } 30 int ans = val[pos]/2 + min(val[pos]/2, len); 31 v.push_back(an[ans+ll-1]); 32 if(ll <= ans+ll-2){ 33 q.push({ll, ans+ll-2}); 34 } 35 if(ans+ll <= rr){ 36 q.push({ans+ll, rr}); 37 } 38 } 39 for(int i = 0 ; i < v.size(); i++){ 40 printf("%d%c", v[i], i == v.size()-1?'\n':' '); 41 } 42 return 0; 43 }

 

转载于:https://www.cnblogs.com/zllwxm123/p/11197420.html