list.remove(list.size()-1);

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]
]

/**
 * 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>>();
		List<Integer> list = new ArrayList<Integer>();
		solve(root, sum, res, list);
		return res;
	}

	private void solve(TreeNode root, int sum, List<List<Integer>> res,
			List<Integer> list) {
		if (root == null) {
			return;
		}
		list.add(root.val);
		sum -= root.val;
		if (root.left == null && root.right == null) {
			if (sum == 0) {
				res.add(new ArrayList<Integer>(list));
			}
		} else {
			if (root.left != null) {
				solve(root.left, sum, res, list);
			}
			if (root.right != null) {
				solve(root.right, sum, res, list);
			}
		}
		list.remove(list.size()-1);
	}
}