Path Sum

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.

• Time: O(N)
• Space: O(1)

public boolean hasPathSum(TreeNode root, int sum) {
    if (root == null) {
        return false;
    }
    if (root.left == null && root.right == null && root.val == sum) {
        return true;
    }
    return hasPathSum(root.left, sum - root.val) 
              || hasPathSum(root.right, sum - root.val);
}

Path Sum II

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

• Time: O(N)
• Space: O(1)

public List<List<Integer>> pathSum(TreeNode root, int sum) {
    List<List<Integer>> ret = new ArrayList<>();
    pathSumRe(ret, new ArrayList<Integer>(), root, sum);
    return ret;
}

private void pathSumRe(List<List<Integer>> ret, List<Integer> path, TreeNode node, int sum) {
    if (node == null) {
        return;
    }
    path.add(node.val);
    if (node.left == null && node.right == null && node.val == sum) {
        ret.add(new ArrayList<>(path));
    }
    pathSumRe(ret, path, node.left, sum - node.val);
    pathSumRe(ret, path, node.right, sum - node.val);
    path.remove(path.size() - 1);
}