leetcode-路径总和 II

给你二叉树的根节点 root 和一个整数目标和 targetSum ，找出所有 从根节点到叶子节点 路径总和等于给定目标和的路径。

叶子节点 是指没有子节点的节点。

示例 1：

输入：root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
输出：[[5,4,11,2],[5,8,4,5]]
示例 2：

输入：root = [1,2,3], targetSum = 5
输出：[]
示例 3：

输入：root = [1,2], targetSum = 0
输出：[]

提示：

树中节点总数在范围 [0, 5000] 内
-1000 <= Node.val <= 1000
-1000 <= targetSum <= 1000

java代码：

/**
 * 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 {
    
    public List<List<Integer>> pathSum(TreeNode root, int targetSum) {
        List<List<Integer>> res = new ArrayList<>();
        huisu(root, targetSum, res, new ArrayList<>(), 0);
        return res;
    }

    private void huisu(TreeNode root, int targetSum, List<List<Integer>> res, ArrayList<Integer> subList, int sum) {
        if (root == null) {
            return;
        }

        sum += root.val;
        subList.add(root.val);
        if (root.left == null && root.right == null) {
            if (sum == targetSum) {
                res.add(new ArrayList<>(subList));
            }
        }

        huisu(root.left, targetSum, res, subList, sum);
        huisu(root.right, targetSum, res, subList, sum);
        subList.remove(subList.size() - 1);

    }
}