0113. Path Sum II

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {

// Time Complexity: O(n^2)
// In the worst case, the tree is a complete binary tree with n nodes. For each node, we traverse all the way down a path till leaf, which takes O(n) time. And there can be n paths in the worst case. So the overall time complexity is O(n^2).

// Space Complexity: O(n)
// The recursion stack depth can go upto n in the worst case. Additionally, the curr list storing the path can contain at most n nodes. So the overall space complexity is O(n).


    public void helper(int target, TreeNode root, List<Integer> curr, List<List<Integer>> res){
        if (root == null) {
            return;
        }
        target -= root.val;
        if (target == 0 && root.left == null && root.right == null) {
            ArrayList<Integer> newCurr = new ArrayList<>(curr);
            newCurr.add(root.val);
            res.add(newCurr);
            return;
        }
        else {
            curr.add(root.val);
            helper(target, root.left, curr, res);
            helper(target, root.right, curr, res);
            curr.remove(curr.size() - 1);
            return;
        }
    }

    public List<List<Integer>> pathSum(TreeNode root, int targetSum) {
        List<List<Integer>> res = new ArrayList<>();
        if (root == null) {
            return res;
        }
        List<Integer> curr = new ArrayList<>();
        helper(targetSum, root, curr, res);
        return res;
    }
}