力扣 113. 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。

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]]

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

输入：root = [1,2], targetSum = 0
输出：[]

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/path-sum-ii
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

class Solution {
    List> res = new ArrayList<>();
    public List> pathSum(TreeNode root, int targetSum) {
        if (root == null) return res;
        linkedList track = new linkedList<>();
        dfs(0, root, targetSum, track);
        return res;
    }

    public void dfs(int num, TreeNode root, int targetSum, linkedList track) {
        //base case
        if (root == null) return;

		//做选择
        track.add(root.val);
        
        num += root.val;
        
		//满足叶子结点、值相等
        if (num == targetSum && root.left == null && root.right == null) {
            res.add(new ArrayList(track));
        }


        dfs(num, root.left, targetSum, track);
        dfs(num, root.right, targetSum, track);

		//撤销选择
        track.removeLast();
    }
}

class Solution {
    public List> res = new ArrayList<>();
    
    public List> pathSum(TreeNode root, int targetSum) {
        linkedList list = new linkedList<>();
        getPathSum(root, targetSum, list);
        
        return res;
    }
    
    public void getPathSum(TreeNode root, int targetSum, linkedList trace) {
        if (root == null) return;
        
        // 是叶节点且满足路径和
        if (root.left == null && root.right == null && targetSum == root.val) {
            res.add(new linkedList<>(trace));
        }
        // 做选择
        trace.add(root.val);
        // 进入下一层
        getPathSum(root.left, targetSum - root.val, trace);
        getPathSum(root.right, targetSum - root.val, trace);
        // 回溯
        trace.removeLast();
    }
}