314. Binary Tree Vertical Order Traversal

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    // HashMap key: Horizontal distance; value: Node
    class Node {
        TreeNode treeNode;
        int hd;
        public Node(TreeNode treeNode, int hd) {
            this.treeNode = treeNode;
            this.hd = hd;
        }
    }
    public List<List<Integer>> verticalOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        if (root == null) {
            return res;
        }
        Map<Integer, List<Integer>> map = new HashMap<>();
        int minHd = 0;
        int maxHd = 0;
        Queue<Node> queue = new LinkedList<>();
        queue.offer(new Node(root, 0));
        while (!queue.isEmpty()) {
            Node cur = queue.poll();
            int hd = cur.hd;
            TreeNode treeNode = cur.treeNode;
            if (!map.containsKey(hd)) {
                map.put(hd, new ArrayList<>());
            }
            map.get(hd).add(treeNode.val);
            if (treeNode.left != null) {
                queue.offer(new Node(treeNode.left, hd - 1));
                minHd = Math.min(minHd, hd - 1);
            }
            if (treeNode.right != null) {
                queue.offer(new Node(treeNode.right, hd + 1));
                maxHd = Math.max(maxHd, hd + 1);
            }
        }
        for (int i = minHd; i <= maxHd; i++) {
            List<Integer> cur = map.get(i);
            res.add(cur);
        }
        return res;
    }
}