浙大数据结构慕课课后题(04-树6 Complete Binary Search Tree)
题目要求：
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.
输入规格：
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.
输出规格：
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.
样例输入：
10
1 2 3 4 5 6 7 8 9 0
样例输出：
6 3 8 1 5 7 9 0 2 4
题解：
思路如注释所示，可通过所有测试点。
#include<bits/stdc++.h>
using namespace std;
int num[1005] = {0} ,ans[1005] = {0};int compare(const void*a,const void*b){ //qsort函数的比较规则 return *(int*)a - *(int*)b;
}int GetLeftLength(int N){ //计算左子树数目int H,X,L;H = log2(N+1);X = N - pow(2, H) + 1;X < pow(2, H-1) ? X = X : X = pow(2, H-1);L = pow(2, H-1) -1 + X;return L;
}void solve(int Left, int Right, int TRoot){int n,L,LeftTRoot,RightTRoot;n = Right - Left + 1;if(n == 0) return; //处理长度为0,则说明已经处理完了L = GetLeftLength(n);ans[TRoot] = num[Left + L];LeftTRoot = TRoot * 2 + 1;RightTRoot = LeftTRoot + 1;solve(Left, Left + L - 1, LeftTRoot); //递归处理左子树 solve(Left + L + 1, Right, RightTRoot); //递归处理右子树
}int main(){int N;cin>>N;for(int i = 0; i < N; i++){cin>>num[i]; }qsort(num, N, sizeof(int), compare);solve(0, N-1, 0);for(int i = 0;i < N; i++){cout<<ans[i];if(i != N-1)cout<<" ";}}