데이터 구조 트 리 예제 Complete Binary Search Tree

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

#include 
#include 
#include 

int compare(const void* a, const void* b);
void Solve(int A[], int T[], int ALeft, int ARight, int TRoot);
int GetLeftLength(int n);

int main()
{
    int N, IsLegal = 0;//N     
    int* A, *T;//        ，                        
    int i;

    QQ: ;
    IsLegal = scanf("%d", &N);
    if(N < 0 || N > 1000 || IsLegal != 1){
        printf("N      \t     
");
        goto QQ;
    }
    T = (int*)malloc(N * sizeof(int));
    A = (int*)malloc(N * sizeof(int));
    if(A == NULL){
        printf("  A         
");
        exit(1);
    }else if(T == NULL){
        printf("  T         
");
        exit(1);
    }
    for(i = 0; i < N; i++){
        scanf("%d", &A[i]);
    }
    qsort(A, N, sizeof(int), compare);//              
    Solve(A, T, 0, N - 1, 0);
    for(i = 0; i < N - 1; i++){
        printf("%d ", T[i]);
    }
    printf("%d
", T[N - 1]);
    free(A);
    free(T);
    return 0;
}

int compare(const void* a, const void* b)
{
    return *(int*)a - *(int*)b;
}

void Solve(int A[], int T[], int ALeft, int ARight, int TRoot)
{//    T，    ，          
    int n, L, LeftRoot, RightRoot;
    n = ARight - ALeft + 1;
    if(n == 0) return;//           
    L = GetLeftLength(n);//  n     ，          
    T[TRoot] = A[ALeft + L];//                 ，         T      
    LeftRoot = TRoot * 2 + 1;//           T     
    RightRoot = LeftRoot + 1;//           T     
    Solve(A, T, ALeft, ALeft + L -1, LeftRoot);
    Solve(A, T, ALeft + L + 1, ARight, RightRoot);
}

int GetLeftLength(int n)
{
    int H, x1, x2, x, L;

    H = floor((log10(n + 1)) / log10(2));
    x1 = n + 1 - pow(2, H);
    x2 = pow(2, H - 1);
    if(x1 < x2){
        x = x1;
    }else{
        x = x2;
    }
    L = x2 - 1 + x;
    return L;
}