LeetCode Path Sum II

原题链接在这里：https://leetcode.com/problems/path-sum-ii/

递归调用，终止条件是当遇到叶子节点时判断sum是否为0，则res加当前ls. 若root.left 不为空，则ls.add(root.left.val)然后递归调用helper, 用完后要remove掉尾节点。右侧相同。

Note: 1. 当res加 ls时一定要res.add(new ArrayList(ls)), 因为list 是 pass by reference, 后面若更改ls, 则已经加到res里的ls也会同时更改。

2. 去掉list尾部就用ls.remove(ls.size()-1).

AC Java:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        if(root == null){
            return res;
        }
        List<Integer> ls = new ArrayList<Integer>();
        ls.add(root.val);
        helper(root,sum-root.val,res,ls);
        return res;
    }
    private void helper(TreeNode root, int sum, List<List<Integer>> res, List<Integer> ls){
        if(root.left == null && root.right == null && sum == 0){
            res.add(new ArrayList(ls)); //error
            return;
        }
        if(root.left != null){
            ls.add(root.left.val);
            helper(root.left,sum-root.left.val,res,ls);
            ls.remove(ls.size()-1);
        }
        if(root.right != null){
            ls.add(root.right.val);
            helper(root.right,sum-root.right.val,res,ls);
            ls.remove(ls.size()-1);
        }
    }
}