【剑指offer】Q25：二叉树中和为某一值的路径

说明：最烦的就是看别人的博客，题解里直接上代码，一行分析都没有。只是这个题。。。

class BTNode():
	def __init__(self, val = -1):
		self.val = val
		self.left = None
		self.right = None

class BTree():
	def __init__(self):
		self.root = None
	'''
	ex
	               1
	              / 
	             2   3
	            /   /
	           4   5

	 treeArray = [1,2,3,4,'#',5]
	'''
	def createTree(self, treeArray):
		self._createTree(treeArray)

	def _createTree(self,treeArray, i = 0):
		if i > len(treeArray) :
			return None
		if treeArray[i] == '#':
			return None

		root = BTNode(int(treeArray[i]))
		if self.root == None:
			self.root = root

		#create left branch
		l = 2*i + 1
		if l < len(treeArray):
			root.left = self._createTree(treeArray, l)

		#create right branch
		r = 2*i + 2
		if r < len(treeArray):
			root.right = self._createTree(treeArray,r)
		return root

    
	def preorder(self, root):
		if root == None:
			return
		print root.val
		self.preorder(root.left)
		self.preorder(root.right)

	def inorder(self, root):
		if root == None:
			return
		self.inorder(root.left)
		print root.val
		self.inorder(root.right)

	def postorder(self, root):
		if root == None:
			return
		self.postorder(root.left)
		self.postorder(root.right)
		print root.val

def Print(path):
	for i in range(len(path)):
		print path[i]
	print "--------------------"

def pathSum(broot, remainder, path):
	if broot == None:
		return
	if remainder < broot.val:
		return
	path.append(broot.val)
	remainder -= broot.val
	if remainder == 0:
		if broot.left == None and broot.right == None:
			Print(path)
		else:
			return
	pathSum(broot.left, remainder, path)
	pathSum(broot.right, remainder,path)
	path = path.pop()



if __name__ == '__main__':
	array = [10,5,12,4,7]
	bt = BTree()
	bt.createTree(array)
	#bt.preorder(bt.root)
	path = []
	remainder = 22
	pathSum(bt.root, remainder, path)