CODE 20: Path Sum II

public ArrayList<ArrayList<Integer>> pathSum(TreeNode root, int sum) {
// Start typing your Java solution below
// DO NOT write main() function
ArrayList<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]
]