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]
]
与Path Sum相比,这道题要求我们输出所有可能的结果,我们同样采用DFS,因为要输出所有可能的结果,我们用回溯法来记录,当找到符合条件的路径就记录下来,然后回溯,继续查找其它的路径。代码如下:
/** * 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>> result = new ArrayList<List<Integer>>(); LinkedList<Integer> list = new LinkedList<Integer>(); if(root == null) return result; getPath(root, 0, sum, result, list); return result; } public void getPath(TreeNode root, int count, int sum, List<List<Integer>> result, LinkedList<Integer> list) { if(root == null) return; count += root.val; list.add(root.val); if(count == sum && root.left == null && root.right == null) { result.add(new LinkedList<Integer>(list)); list.removeLast(); return; } getPath(root.left, count, sum, result, list); getPath(root.right, count, sum, result, list); list.removeLast(); } }