CODE 20: Path Sum II

public ArrayList<ArrayList<Integer>> pathSum(TreeNode root, int sum) {// Start typing your Java solution below// DO NOT write main() functionArrayList<ArrayList<Integer>> paths = new ArrayList<ArrayList<Integer>>();if (null == root) {return paths;}dfs(paths, new ArrayList<Integer>(), root, 0, sum);return paths;}private void dfs(ArrayList<ArrayList<Integer>> paths,ArrayList<Integer> path, TreeNode root, int current, int sum) {if (root.left == null && root.right == null) {if (current + root.val == sum) {path.add(root.val);ArrayList<Integer> pathCopy = new ArrayList<Integer>();pathCopy.addAll(path);paths.add(pathCopy);path.remove(path.size() - 1);}}path.add(root.val);current += root.val;if (root.left != null) {dfs(paths, path, root.left, current, sum);}if (root.right != null) {dfs(paths, path, root.right, current, sum);}path.remove(path.size() - 1);}

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]]