113. 路径总和 II

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def pathSum(self, root: TreeNode, sum: int) -> List[List[int]]:
        def dfs(root, sum):
            if root is None:
                return
            path.append(root.val)
            if root.val == sum and root.left is None and root.right is None:
                res.append(path.copy())
            dfs(root.left, sum - root.val)
            dfs(root.right, sum - root.val)
            path.pop()
        if not root:
            return []
        res = []
        path = []
        dfs(root, sum)
        return res

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    private List<List<Integer>> res;
    private List<Integer> path;

    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        if (root == null) return Collections.emptyList();
        res = new ArrayList<>();
        path = new ArrayList<>();
        dfs(root, sum);
        return res;
    }

    private void dfs(TreeNode root, int sum) {
        if (root == null) return;
        path.add(root.val);
        if (root.val == sum && root.left == null && root.right == null) {
            res.add(new ArrayList<>(path));
        }
        dfs(root.left, sum - root.val);
        dfs(root.right, sum - root.val);
        path.remove(path.size() - 1);
    }
}