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/*

Author: King, higuige@gmail.com

Date: Oct 7, 2014

Problem: Path Sum 2

Difficulty: easy

Source: https://oj.leetcode.com/problems/path-sum-ii/

Notes:

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

For example:

Given the below binary tree and sum = 22,

5

/ \

4 8

/ / \

11 13 4

/ \ / \

7 2 5 1

return

[

[5,4,11,2],

[5,8,4,5]

]

Solution: DFS.

*/

/**

* Definition for binary tree

* public class TreeNode {

* int val;

* TreeNode left;

* TreeNode right;

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

* }

*/

public class Solution {

public List<List<Integer>> pathSum(TreeNode root, int sum) {

List<List<Integer>> res = new ArrayList<List<Integer>>();

ArrayList<Integer> path = new ArrayList<Integer>();

pathSumRe(root, sum, res, path);

return res;

}

public void pathSumRe(TreeNode root, int sum, List<List<Integer>> res, ArrayList<Integer> path)

{

if (root == null) return;

path.add(root.val);

if (root.left == null && root.right == null && root.val == sum){

ArrayList<Integer> tmp = new ArrayList<Integer>(path);

res.add(tmp);

}

pathSumRe(root.left, sum - root.val, res, path);

pathSumRe(root.right, sum - root.val, res, path);

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

}

}