113. Path Sum II (root-to-leaf, find path for target)

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

Note: A leaf is a node with no children.

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]
]

Thoughts:

1. Standard DFS backTracking

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        List<List<Integer>> res = new ArrayList<>();
        List<Integer> path = new ArrayList<>();
        dfs(res, path, root, sum);
        return res;
    }
    private void dfs(List<List<Integer>> res, List<Integer>path, TreeNode root, int sum){
        if(root == null) return;
        path.add(root.val);
        if(root.left == null && root.right == null){
            if(sum == root.val) res.add(new ArrayList<>(path));
            path.remove(path.size() - 1);
            return;
        }
        dfs(res, path, root.left, sum - root.val);
        dfs(res, path, root.right, sum - root.val);
        path.remove(path.size() - 1);
    }
}