面试题34. 二叉树中和为某一值的路径

/**
 * 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 {
    LinkedList<List<Integer>> res = new LinkedList<>();
    LinkedList<Integer> path = new LinkedList<>();

    //List 不是数据结构，ArrayList 是数组，LinkedList 是双向链表，
    //List 是接口，接口可以作为返回值
    //但是上面因为是接口，所以不能 List<List<Integer>> res = new List<>();
    public List<List<Integer>> pathSum(TreeNode root, int target) {
        recur(root, target);
        return res;
    }

    
    void recur(TreeNode root, int tar){
        //若节点 root 为空，则直接返回。
        if(root == null) return;
        //更新目标值
        tar -= root.val;
        //添加当前路径
        path.add(root.val);
        //找到了一条路径
        if(tar == 0 && root.left == null && root.right == null)
            res.add(new LinkedList(path));
        //继续向两边递归
        recur(root.left, tar);
        recur(root.right ,tar);
        //马上这一层递归就要结束了，去除这层递归的路径
        path.removeLast();
    }
}