#leetcode#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

return

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

分析： DFS backtracking... ebay面过一个题， print the path from root to a given node， 早练练这题的话那个题也不会挂了，之所以会挂主要是对Java Pass by Value的理解不到位，还有就是可以设一个全局变量boolean flag，来设置找到given node之后就不继续做recursion了

/**
* 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>> res = new ArrayList<List<Integer>>();
if(root == null){
return res;
}
helper(res, root, sum, new ArrayList<Integer>());
return res;
}

private void helper(List<List<Integer>> res, TreeNode root, int sum, List<Integer> item){
if(root == null){
return;
}
if(root.left == null && root.right == null && root.val == sum){
item.add(root.val);
res.add(new ArrayList<Integer>(item));
item.remove(item.size() - 1);
return;
}

item.add(root.val);
helper(res, root.left, sum - root.val, item);
helper(res, root.right, sum - root.val, item);
item.remove(item.size() - 1);
}
}