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

解法1，DFS：

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def pathSum(self, root, sum):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: int
        # java solution
        public class Solution {
            public int pathSum(TreeNode root, int sum) {
                if (root == null) return 0;
                return pathSumFrom(root, sum) + pathSum(root.left, sum) + pathSum(root.right, sum);
            }

            private int pathSumFrom(TreeNode node, int sum) {
                if (node == null) return 0;
                return (node.val == sum ? 1 : 0) 
                    + pathSumFrom(node.left, sum - node.val) + pathSumFrom(node.right, sum - node.val);
            }
        }
        """
        def dfs(node, target):      
            if not node: return 0 
            return int(node.val == target)+dfs(node.left, target-node.val)+dfs(node.right, target-node.val)
        
        if not root: return 0
        return dfs(root, sum)+self.pathSum(root.left, sum)+self.pathSum(root.right, sum)

尤其註意是dfs(root, sum)+self.pathSum(root.left, sum)+self.pathSum(root.right, sum) 而不是dfs+dfs+dfs！容易出錯！

另外解法就是前綴樹的解法，前綴樹記錄sum，比如：

root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8

      10
     /      5   -3
   / \    \
  3   2   11
 / \   \
3  -2   1

到節點3的前綴sum就是{10:1, 15:1, 18:1}，則包括節點3的滿足sum=8的路徑就是{10,5,3}這條通路減去{10}這條通路，兩個前綴相減，更深的前綴-淺的前綴就是中間路路徑

代碼：

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def pathSum(self, root, sum):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: int        
        """
        def traverse(node, total, pre_sum):
            if not node:
                return 0
            total += node.val
            ans = pre_sum.get(total-sum, 0)
            pre_sum[total] = pre_sum.get(total, 0)+1
            ans += traverse(node.left, total, pre_sum) + traverse(node.right, total, pre_sum)
            pre_sum[total] -= 1
            return ans
        
        return traverse(root, 0, {0:1})        

leetcode 437. Path Sum III