Path Sum II | Leetcode

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: Maintain a current list of node values as you are recursing down each path. Subtract the node's value from sum each time. When you reach a leaf node (node.left == node.right == null), check if the sum is 0. If it were 0, add it to the list of answers.

/**
 * 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>> answer = new ArrayList<>();
        if (root == null) return answer;
        pathSum(answer, new ArrayList<Integer>(), root, sum);
        return answer;
    }
    
    private void pathSum(List<List<Integer>> answer, List<Integer> list, TreeNode node, int sum) {
        sum -= node.val;
        list.add(node.val);
        if (node.left == null && node.right == null) {
            if (sum == 0) answer.add(new ArrayList<Integer>(list));
        } else {
            if (node.left != null) pathSum(answer, list, node.left, sum);
            if (node.right != null) pathSum(answer, list, node.right, sum);
        }
        list.remove(list.size() - 1);
    }
}