Path Sum I && II

I: Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

For example:
Given the below binary tree and sum = 22,

              5
             / \
            4   8
           /   / \
          11  13  4
         /  \      \
        7    2      1

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

DFS: 

public class Solution {
    //Time: O(n)
    public boolean hasPathSum(TreeNode root, int sum) {
        if (root == null) {
            return false;
        }
        return checkPath(root, root.val, sum);
    }
    
    private boolean checkPath(TreeNode root, int cur, int sum) {
        if (root.left == null && root.right == null) {
            if (cur == sum) {
                return true;
            }
            return false;
        }
        
        boolean left = false;
        boolean right = false;;
        if (root.left != null) {
            left = checkPath(root.left, cur + root.left.val, sum);
        }
        if (root.right != null) {
            right = checkPath(root.right, cur + root.right.val, sum);
        }
        return left || right;
    }
}

BFS: 

public class Solution {
    //Time: O(n)
    public boolean hasPathSum(TreeNode root, int sum) {
        if (root == null) {
            return false;
        }
        
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        Queue<Integer> queueSum = new LinkedList<Integer>();
        queue.offer(root);
        queueSum.offer(root.val);
        while (!queue.isEmpty()) {
            TreeNode cur = queue.poll();
            int cursum = queueSum.poll();
            if (cur.left != null) {
                queue.offer(cur.left);
                queueSum.offer(cursum + cur.left.val);
            }
            if (cur.right != null) {
                queue.offer(cur.right);
                queueSum.offer(cursum + cur.right.val);
            }
            if (cur.left == null && cur.right == null && cursum == sum) {
                return true;
            }
        }
        return false;
    }
}

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

DFS: 

public class Solution {
    public ArrayList<ArrayList<Integer>> pathSum(TreeNode root, int sum) {
        ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
        if (root == null) {
            return res;
        }
        ArrayList<Integer> tmp = new ArrayList<Integer>();
        path(res, tmp, root, sum);
        return res;
    }
    
    private void path(ArrayList<ArrayList<Integer>> res, ArrayList<Integer> tmp,
                      TreeNode root, int sum) {
        if (root.left == null && root.right == null) {
            tmp.add(root.val);
            if (sum == root.val) {
                res.add(new ArrayList(tmp));
            }
            tmp.remove(tmp.size() - 1);
            return;
        }   
        
        tmp.add(root.val);
        if (root.left != null) {
            path(res, tmp, root.left, sum - root.val);
        }
        if (root.right != null) {
            path(res, tmp, root.right, sum - root.val);
        }
        tmp.remove(tmp.size() - 1);
    }
}