剑指Offer22 判断数组是否为某二叉搜索树的后序遍历

/*************************************************************************
    > File Name: 22_SequenceOfBST.cpp
    > Author: Juntaran
    > Mail: JuntaranMail@gmail.com
    > Created Time: 2016年08月30日 星期二 20时34分33秒
 ************************************************************************/

#include <stdio.h>
#include <stdlib.h>

// 判断一个数组是不是某个二叉搜索树的后序遍历
bool isSequenceOfBST(int* sequence, int length)
{
    if (sequence==NULL || length<=0)
        return false;
    
    int root = sequence[length-1];
    
    int i, j;
    
    // 二叉搜索树的左子树都小于根
    for (i = 0; i < length-1; ++i)
    {
        if (sequence[i] > root)
            break;
    }
    
    // 二叉搜索树的右子树都大于根
    for (j = i; j < length-1; ++j)
    {
        if (sequence[j] < root)
            return false;
    }
    
    // 递归判断左子树是不是二叉搜索树
    bool left = true;
    if (i > 0)
        left = isSequenceOfBST(sequence, i);
    
    // 递归判断右子树是不是二叉搜索树
    bool right = true;
    if (i < length-1)
        right = isSequenceOfBST(sequence+i, length-i-1);
    
    return (left && right);
}

int main()
{
    int sequence1[] = {5,7,6,9,11,10,8};
    int sequence2[] = {7,4,5,6};
    bool ret1 = isSequenceOfBST(sequence1, 7);
    bool ret2 = isSequenceOfBST(sequence2, 4);
    if (ret1 == true)
        printf("ret1 is true
");
    else
        printf("ret1 is false
");
    
    if (ret2 == true)
        printf("ret2 is true
");
    else
        printf("ret2 is false
");
    
    return 0;
}