问题描述：
You are given a binary tree in which each node contains an integer value.
Find the number of paths that sum to a given value.
The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes).
The tree has no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000.
Example:
root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8
10
/
5 -3
/
3 2 11
/
3 -2 1
Return 3. The paths that sum to 8 are:
1. 5 -> 3
2. 5 -> 2 -> 1
3. -3 -> 11
思路：
仍然是深度优先遍历的原则，该方法需要多做题多巩固啊！！占了我一整天时间的一道题。
从头到尾遍历每一个节点，记录从根节点到每一个节点的sum。
通过查找当前节点和目标sum之间的差值，进行相应剪枝动作
当退出当前层次时候，需要减去当前层次计算出来的和，故每次return之前都有一个减一的动作
代码：
1 # Definition for a binary tree node. 2 # class TreeNode: 3 # def __init__(self, x): 4 # self.val = x 5 # self.left = None 6 # self.right = None 7 8 class Solution: 9 def pathSum(self, root: TreeNode, sum: int) -> int: 10 if root == None : return 0 11 prefix = {0:1} 12 return self.Calc(root,0,sum,prefix) 13 14 15 def Calc(self,root,cursum,target,prefix): 16 if root == None : return 0 17 cursum += root.val 18 res = prefix.get(cursum - target,0) 19 prefix[cursum] = prefix.get(cursum,0) + 1 20 res += self.Calc(root.left,cursum,target,prefix) + self.Calc(root.right,cursum,target,prefix) 21 prefix[cursum] = prefix.get(cursum) - 1 22 return res