113. Path Sum II

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

Note: A leaf is a node with no children.

Example:

Given the below binary tree and sum = 22,

      5
     / \
    4   8
   /   / \
  11  13  4
 /  \    / \
7    2  5   1

Return:

[
   [5,4,11,2],
   [5,8,4,5]
]

 1 class Solution {
 2     public List<List<Integer>> pathSum(TreeNode root, int sum) {
 3         List<List<Integer>> res = new ArrayList();
 4         List<Integer> cur = new ArrayList();
 5         help(root, sum, res, cur);
 6         return res;
 7     }
 8     private void help(TreeNode root, int sum, List<List<Integer>> res,List<Integer> cur){
 9         if(root == null) return;
10         cur.add(root.val);
11         //到达叶子节点
12         if(root.left == null && root.right == null){
13             if(sum == root.val) res.add(new ArrayList(cur));//务必new一个arraylist，否则会修改之前的
14         }
15         //继续递归
16         help(root.left, sum - root.val, res, cur);
17         help(root.right, sum - root.val, res, cur);
18         //表明无论成功与否，此node的使命已经完成，remove以为后续节点留空间
19         cur.remove(cur.size() - 1);
20     }
21 }

和path sum一方法类似，都是减到最后比较root和剩下的值，相等就说明这一组成功，区别是要用一个current的list存放cur结果，如果此数已经被计算过，在最后要进行删除。

转载于:https://www.cnblogs.com/wentiliangkaihua/p/11456997.html