【剑指offer】面试题24：二叉搜索树的兴许前序遍历序列

分析：

前序： 根 左 右

后序： 左 由 根

二叉搜索树： 左 < 根 < 右

那么这就非常明显了。

def ifpost(postArray, start, end):
	#one or "" is true
	if(end - start <= 1):
		return True
	i = start
	#left branch whose value < root
	while i <= end and postArray[i] < postArray[end]:
		i += 1
	#partion of left and right branch
	part = i
	#right branch whose value > root
	while i <= end and postArray[i] > postArray[end]:
		i += 1
	#not all right part > root
	if(i < end):
		return False
	return ifpost(postArray, start, part - 1) and ifpost(postArray, part, end - 1)

def ifpreorder(preArray, start, end):
	if(end - start <= 1):
		return True
	i = end
	while(i >= start and preArray[i] > preArray[start]):
		i -= 1
	partition = i
	while(i >= start and preArray[i] < preArray[start]):
		i -= 1
	if(i > start):
		return False
	return ifpreorder(preArray, start + 1, partition) and \
	 ifpreorder(preArray, partition + 1, end)