04-树6 Complete Binary Search Tree(30 分)

xiaoxiao2021-02-28  5

题目来源:中国大学MOOC-陈越、何钦铭-数据结构-2018春 作者: 陈越 单位: 浙江大学 问题描述: 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

解答: 存储数据的结构均选择数组,因为完全二叉树用数组更好做(可直接利用性质)。 首先,因为搜索二叉树左子节点<根<右子节点,发现若给n个数排好序,则这个序列是完全搜索二叉树的中序遍历。要建立这棵树,光知道中序遍历是不够的,因此引入完全二叉树的性质,就是给每个节点按1-N编号,则设根节点为i,左节点为2i,右子节点为2i+1,这样就可以从根节点1开始中序遍历完全二叉搜索树,在遍历的同时建树。中序遍历是访问节点时cout,那么现在需要建树,我们需要给这个节点赋一个值,这个值就是排好序的n个数依次取出即可。pos变量就是模拟现在取到哪个数了,因此要设置为全局。 注意递归建树的return条件是发现根的值大于N时返回。然后因为完全二叉树编号即是按层从上到下从左到右编号,所以输出直接按序输出Tree数组里的N个数即可。

#include <iostream> #include <algorithm> using namespace std; const int maxn=1001; int N; int input[maxn],tree[maxn],pos=0; void inputData() { cin>>N; for(int i=0;i<N;i++) { cin>>input[i]; } } void buildTree(int root) { int leftChild=2*root,rightChild=2*root+1; if(root>N) return; buildTree(leftChild); tree[root]=input[pos++]; buildTree(rightChild); } void output() { for(int i=1;i<=N;i++) { if(i!=N) cout<<tree[i]<<" "; else cout<<tree[i]; } } int main() { inputData(); sort(input,input+N); buildTree(1); output(); return 0; }
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