04-树6 Complete Binary Search Tree

　　本题要求根据输入实现一棵完美二叉搜索树并层序遍历输出。

　　以下方法没有构造出树再进行遍历，而是利用完美二叉树中各结点位置的规律逐步得出要求的序列，以数组形式存放最后直接输出。

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#define MAXSIZE 1000

int comp(const void *a, const void *b);
void Func(int* inSeq, int* LevTrSeq, int LeftIndex, int RightIndex, int RootIndex);
int GetLeftNum(int num);
int Min(int a, int b);

int main(int argc, char const *argv[])
{
    int N, i;
    int inSeq[MAXSIZE], LevTrSeq[MAXSIZE];
    scanf("%d", &N);
    for (i = 0; i < N; i++){
        scanf("%d", &inSeq[i]);
    }
    qsort(inSeq, N, sizeof(int), comp);
    int LeftIndex = 0, RightIndex = N - 1, RootIndex = 0; 
    Func(inSeq, LevTrSeq, LeftIndex, RightIndex, RootIndex);
    printf("%d", LevTrSeq[0]);
    for (i = 1; i < N; i++){
        printf(" %d", LevTrSeq[i]);
    }
    printf("\n");
    return 0;
}

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

void Func(int* inSeq, int* LevTrSeq, int LeftIndex, int RightIndex, int RootIndex)
{
    int num;  //number of elements in this sequence
    num = RightIndex - LeftIndex + 1;
    if(!num)
        return;
    int LeftNum, LeftRootIndex, RightRootIndex;  //number of nodes in left subtree; substree roots' index
    LeftNum = GetLeftNum(num);
    LevTrSeq[RootIndex] = inSeq[LeftIndex + LeftNum];
    LeftRootIndex = 2 * RootIndex + 1;  //完全二叉树每层最左节点的下标规律如此
    RightRootIndex = LeftRootIndex + 1;  //因为是层序遍历，输出序列中右子树根节点紧挨着左子树根节点
    Func(inSeq, LevTrSeq, LeftIndex, LeftIndex + LeftNum - 1, LeftRootIndex);
    Func(inSeq, LevTrSeq, LeftIndex + LeftNum + 1, RightIndex, RightRootIndex);
    return;
}

int GetLeftNum(int num)
{
    int h = 0;  //tree height
    int i, LowestNum;  //LowestNum is the nodes number of the lowest level
    int LeftNum;
    int LastLevelNum = 1;  //each level nodes number
    int Num = num;  //底下还要用到num，不能直接对其进行操作
    while(Num > 1){
        Num /= 2;
        h++;
    }
    for(i = 0; i < h - 1; i++){
        LastLevelNum *= 2;
    }
    //这里的操作务必是上一层全部节点数乘2，不能先乘到最后一层再用除以2操作。
    //容易除成负数最后负数返回导致数组下标为负出错
    LowestNum = num - LastLevelNum * 2 + 1;
    LeftNum = Min(LastLevelNum, LowestNum) + LastLevelNum - 1;
    return LeftNum;
}

int Min(int a, int b)
{
    return (a < b) ? a : b;
}