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]
]

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();
    }
}