1064. Complete Binary Search Tree (30)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 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

中序遍历完全搜索二叉树 必定得到一个有序的序列,所有先建立一个完全二叉树,中序遍历得到地址数组,然后按照数据从小到大依次填入即可.

#include
    
     #include
     
      #include
      
       #include
       
        #include
        
         #include
         #include
          
           #include
           
            using namespace std;struct Node{ int data; Node *left,*right; Node():data(0),left(NULL),right(NULL){};};void Inorder(Node * T,vector
            
             &ans){ if(T->left) Inorder(T->left, ans); if(T) ans.push_back(T); if(T->right) Inorder(T->right, ans);}int main(){ int n,count=0; cin>>n; vector
             
              nums(n); for (int i=0; i
              
               >nums[i]; } if(n==1) { printf("%d\n",nums[0]); return 0; } sort(nums.begin(), nums.end()); Node * root =NULL; queue
               
                q; root = new Node; count++; if(root) q.push(root); while (!q.empty()) {//建立完全二叉树 Node * p = q.front(); q.pop(); p->left = new Node; count ++; if(count==n) break; q.push(p->left); p->right = new Node; count++; if(count==n) break; q.push(p->right); } vector
                
                 ans; Inorder(root, ans); for(int i=0;i
                 
                  data = nums[i]; } queue
                  
                    level; if(root) level.push(root); vector
                   
                    o; while (!level.empty()) { Node * p = level.front(); level.pop(); o.push_back(p->data); if(p->left)level.push(p->left); if(p->right)level.push(p->right); } for (int i=0; i

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