04-树6 Complete Binary Search Tree (30分)---AI那点小事

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; }

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