给定BST先序遍历序列，不构造BST的情况下判断BST是否每个node都只有一个child

// 给定BST先序遍历序列_判断BST是否每个node都只有一个child.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"

bool isOnlyHaveOneChild(int A[], int len)
{
for (int i = 0; i < len-1; i++)
{
if (A[i] < A[i+1])
{
for (int j = i+1; j < len; j++)
{
if (A[j] < A[i])
{
return false;
}
}
}
else
{
for (int k = i+1; k < len; k++)
{
if (A[k] > A[i])
{
return false;
}
}
}
}

return true;
}

int _tmain(int argc, _TCHAR* argv[])
{

int A1[] = {16,10,8,9,29,27,22};
int A2[] = {10,19,17,14,15,16};

bool a1 =  isOnlyHaveOneChild(A1, 7);
bool a2 =  isOnlyHaveOneChild(A2, 6);

printf("%d, %d\n", a1,a2);

return 0;
}

bool isOnlyHaveOneChild2(int A[], int len)
{
if (A == NULL || len == 0)
return false;

if (len == 1)
return true;

int MinNum;
int MaxNum;

MinNum = MaxNum = A[len-1];

for (int idx = len-2; idx >=0; idx--)
{
if (A[idx] > MinNum && A[idx] < MaxNum)
return false;

if (A[idx] > MaxNum)
MaxNum = A[idx];
else if (A[idx] < MinNum)
MinNum = A[idx];
}
return true;
}