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113. Path Sum II

时间:2018-01-29 23:07:59      阅读:206      评论:0      收藏:0      [点我收藏+]
Given a binary tree and a sum, find all root-to-leaf paths where each path‘s sum equals the given sum.
For 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     List<List<Integer>> res = new ArrayList<>();
 3     public List<List<Integer>> pathSum(TreeNode root, int sum) {
 4         List<Integer> curres = new ArrayList<Integer>();
 5         help(root,0,sum,curres);
 6         return res;
 7     }
 8     private void  help(TreeNode root,int sum ,int target,List<Integer> curres){
 9         if(root == null) return ;
10         curres.add(root.val);
11         if(root.left==null && root.right==null && sum+root.val==target) 
12             res.add(new ArrayList(curres));
13         help(root.left,sum+root.val,target,curres);
14         help(root.right,sum+root.val,target,curres);
15         curres.remove(curres.size()-1);
16     }
17 }

 

113. Path Sum II

原文:https://www.cnblogs.com/zle1992/p/8379706.html

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