二叉树中和为某一值的路径

解题：前序遍历加上筛选

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    LinkedList<Integer> list = new LinkedList<>();
    LinkedList<List<Integer>> res = new LinkedList<>();
    public List<List<Integer>> pathSum(TreeNode root, int sum) {
    helper(root,sum,list);
    return res;
    }
    //先序遍历
    public void helper(TreeNode root,int sum,LinkedList<Integer> list){
        if(root == null) return ;
         list.add(root.val);
         sum -= root.val;
        if(sum == 0 && root.left == null && root.right ==null){
            res.add(new LinkedList(list));
          //  return ;
        }
        helper(root.left,sum,list);
        helper(root.right,sum,list); 
        list.removeLast();
    }
}

标签：TreeNode,LinkedList,helper,sum,路径,list,二叉树,root,一值
来源： https://www.cnblogs.com/cstdio1/p/13368525.html