Path Sum II

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Problem:

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]]

Solution: will add idea later. But this method wastes a lot spaces.

Code:

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();    public ArrayList<ArrayList<Integer>> pathSum(TreeNode root, int sum) {        if (root == null) return result;        helper(root, sum, new ArrayList<Integer>());        return result;            }    public void helper(TreeNode node, int sum, ArrayList<Integer> list) {        if (node == null) return;        list.add(node.val);        if (node.left == null && node.right == null) {            if (sum - node.val == 0) {                ArrayList<Integer> tmp = new ArrayList<Integer>(list);                result.add(tmp);                            }            return;        }        ArrayList<Integer> listLeft = new ArrayList<Integer>(list);        ArrayList<Integer> listRight = new ArrayList<Integer>(list);        helper(node.left, sum - node.val, listLeft);        helper(node.right, sum - node.val, listRight);    }}

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