04-树8. Complete Binary Search Tree (30)

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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.

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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

代码如下：

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#include<iostream>
#include<algorithm>
#include<vector>
#include<cmath>
using namespace std;

int GetLeftTreeNum(int totalNum)
{
    int level = (int)log2(totalNum);
    int bottomNum = min(totalNum + 1 - pow(2, level), pow(2, level - 1));
    return pow(2, level - 1) - 1 + bottomNum;
}

void solve(int beg, int end, int insertPos, vector<int> &nodeVec, vector<int> &inorderVec)
{
    int totalNum = end - beg + 1;
    if(0 == totalNum)
    {
        return;
    }
    int leftTreeNum = GetLeftTreeNum(totalNum);
    inorderVec[insertPos] = nodeVec[beg + leftTreeNum];
    int leftPos = insertPos * 2 + 1;
    int rightPos = leftPos + 1;
    solve(beg, beg + leftTreeNum - 1, leftPos, nodeVec, inorderVec);
    solve(beg + leftTreeNum + 1, end, rightPos, nodeVec, inorderVec);

}

int main()
{
    int N = 0;  //the number of nodes
    cin >> N;

    vector<int> nodesVec;
    vector<int> inorderVec(N, 0);
    for(int i = 0; i < N; ++i)
    {
        int val = 0;
        cin >> val;
        nodesVec.push_back(val);
    }
    sort(nodesVec.begin(),nodesVec.end(),less<int>());

    solve(0, N - 1, 0, nodesVec, inorderVec);

    for(int i = 0; i < inorderVec.size(); ++i)
    {
        cout << inorderVec[i] << (i == inorderVec.size()-1 ? "\n" : " ");
    }

    return 0;
}

第一次没写出来，后来放弃了，直到看到图的时候，老师讲解之后才写出来的，真是惭愧啊！还是太渣，得继续努力！