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_113_PathSumII.java
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_113_PathSumII.java
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package leetcode_1To300;
import java.util.ArrayList;
import java.util.List;
/**
* 本代码来自 Cspiration,由 @Cspiration 提供
* 题目来源:http://leetcode.com
* - Cspiration 致力于在 CS 领域内帮助中国人找到工作,让更多海外国人受益
* - 现有课程:Leetcode Java 版本视频讲解(1-900题)(上)(中)(下)三部
* - 算法基础知识(上)(下)两部;题型技巧讲解(上)(下)两部
* - 节省刷题时间,效率提高2-3倍,初学者轻松一天10题,入门者轻松一天20题
* - 讲师:Edward Shi
* - 官方网站:https://cspiration.com
* - 版权所有,转发请注明出处
*/
public class _113_PathSumII {
/**
* 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.
For example:
Given the below binary tree and sum = 22,
5
/ \
4 8
/ / \
11 13 4
/ \ / \
7 2 5 1
[
[5,4,11,2],
[5,8,4,5]
]
time : O(n);
space : O(n);
* @param root
* @param sum
* @return
*/
public static List<List<Integer>> pathSum(TreeNode root, int sum) {
List<List<Integer>> res = new ArrayList<>();
if (root == null) return res;
helper(res, new ArrayList<>(), root, sum);
return res;
}
public static void helper(List<List<Integer>> res, List<Integer> list, TreeNode root, int sum) {
if (root == null) return;
list.add(root.val);
if (root.left == null && root.right == null) {
if (sum == root.val) {
res.add(new ArrayList<>(list));
}
}
helper(res, list, root.left, sum - root.val);
helper(res, list, root.right, sum - root.val);
list.remove(list.size() - 1);
}
}