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] ]

### Approach 1:

/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { private List<List<Integer>> resultList = new ArrayList<List<Integer>>(); public void pathSumInner(TreeNode root, int sum, Stack<Integer>path) { path.push(root.val); if(root.left == null && root.right == null) if(sum == root.val) resultList.add(new ArrayList<Integer>(path)); if(root.left!=null) pathSumInner(root.left, sum-root.val, path); if(root.right!=null)pathSumInner(root.right, sum-root.val, path); path.pop(); } public List<List<Integer>> pathSum(TreeNode root, int sum) { if(root==null) return resultList; Stack<Integer> path = new Stack<Integer>(); pathSumInner(root, sum, path); return resultList; } }

### Approach 2:

/** * Definition for a binary tree node. * 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>> result = new ArrayList<>(); helper(root, sum, result, new ArrayList<>()); return result; } private void helper(TreeNode root, int sum, List<List<Integer>> result, List<Integer> list) { if (root == null) { return; } list.add(root.val); if (root.left == null && root.right == null && root.val == sum) { result.add(new ArrayList<>(list)); list.remove(list.size()-1); return; } helper(root.left, sum - root.val, result, list); helper(root.right, sum - root.val, result, list); list.remove(list.size()-1); } }