https://leetcode.com/problems/path-sum-iii/description/

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

解题：这题要求二叉树有几条path的值的和等于k,path起止可以不是root或leaf，但是需要从上到下。这里就可以用到前面一题的preSum解法，用一个哈希表来维护当前sum，用sum-k来检查从当前节点有没有到之前的莫一个节点的差为k，如果能找到那个节点说明这一段是个valid path。

class Solution {
    public int pathSum(TreeNode root, int sum) {
        HashMap<Integer, Integer> preSum = new HashMap();
        preSum.put(0,1);
        helper(root, 0, sum, preSum);
        return count;
    }
    int count = 0;
    public void helper (TreeNode root, int currSum, int target, HashMap<Integer, Integer> preSum) {
        if (root == null) return ;

        currSum += root.val;
        if (preSum.containsKey(currSum - target)) {
            count += preSum.get(currSum - target);
        }
        preSum.put(currSum, preSum.getOrDefault(currSum, 0) + 1);

        helper(root.left, currSum, target, preSum);
        helper(root.right, currSum, target, preSum);
        preSum.put(currSum, preSum.get(currSum) - 1);
    }
}