113. Path Sum II
题目：
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] ]
链接： http://leetcode.com/problems/path-sum-ii/
5/9/2017
算法班
3ms, 52%
DFS + backtracking
注意的问题：
ArrayList.removeRange(start, end)是protected的，所以第21，26行不能调用此函数并尝试不做第15行的remove。
可以把第13，19，24行的list.add()放到第12行之前来refactor
1 public class Solution { 2 public List<List<Integer>> pathSum(TreeNode root, int sum) { 3 List<List<Integer>> ret = new ArrayList<>(); 4 if (root == null) return ret; 5 ArrayList<Integer> list = new ArrayList<Integer>(); 6 pathSum(root, ret, sum, list); 7 return ret; 8 } 9 10 private void pathSum(TreeNode root, List<List<Integer>> ret, int sum, ArrayList<Integer> list) { 11 if (root == null) return; 12 if (root.left == null && root.right == null && root.val == sum) { 13 list.add(root.val); 14 ret.add(new ArrayList<Integer>(list)); 15 list.remove(list.size() - 1); 16 return; 17 } 18 if (root.left != null) { 19 list.add(root.val); 20 pathSum(root.left, ret, sum - root.val, list); 21 list.remove(list.size() - 1); 22 } 23 if (root.right != null) { 24 list.add(root.val); 25 pathSum(root.right, ret, sum - root.val, list); 26 list.remove(list.size() - 1); 27 } 28 return; 29 } 30 }
有人尝试做iterative的算法：
https://discuss.leetcode.com/topic/31698/java-solution-iterative-and-recursive
也有BFS + queue
https://discuss.leetcode.com/topic/18444/python-solutions-recursively-bfs-queue-dfs-stack
更多讨论：
