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/**

* 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

*

*/

/**

* Definition for a binary tree node.

* public class TreeNode {

* int val;

* TreeNode left;

* TreeNode right;

* TreeNode(int x) { val = x; }

* }

*/

public class PathSumIII437 {

public int pathSum(TreeNode root, int sum) {

int[] res = new int[1];

pathSum(root, sum, res, new ArrayList<>());

return res[0];

}

private void pathSum(TreeNode root, int target, int[] res, List<Integer> path) {

if (root == null) return;

int last = path.isEmpty() ? root.val : path.get(path.size() - 1) + root.val;

if (last == target) res[0]++;

for (int s: path) {

if (last - s == target) {

res[0]++;

}

}

path.add(last);

pathSum(root.left, target, res, path);

pathSum(root.right, target, res, path);

path.remove(path.size() - 1);

}

/**

* https://leetcode.com/problems/path-sum-iii/discuss/91878/17-ms-O(n)-java-Prefix-sum-method

*/

public int pathSum2(TreeNode root, int sum) {

HashMap<Integer, Integer> preSum = new HashMap();

preSum.put(0,1);

return helper(root, 0, sum, preSum);

}

public int helper(TreeNode root, int currSum, int target, HashMap<Integer, Integer> preSum) {

if (root == null) {

return 0;

}

currSum += root.val;

int res = preSum.getOrDefault(currSum - target, 0);

preSum.put(currSum, preSum.getOrDefault(currSum, 0) + 1);

res += helper(root.left, currSum, target, preSum) + helper(root.right, currSum, target, preSum);

preSum.put(currSum, preSum.get(currSum) - 1);

return res;

}

}