Given a binary tree, return the vertical order traversal of its nodes' values. (ie, from top to bottom, column by column).

If two nodes are in the same row and column, the order should be from left to right.

Examples 1:

Input: [3,9,20,null,null,15,7]

   3
  /\
 /  \
 9  20
    /\
   /  \
  15   7 

Output:

[
  [9],
  [3,15],
  [20],
  [7]
]

Examples 2:

Input: [3,9,8,4,0,1,7]

     3
    /\
   /  \
   9   8
  /\  /\
 /  \/  \
 4  01   7 

Output:

[
  [4],
  [9],
  [3,0,1],
  [8],
  [7]
]

Examples 3:

Input: [3,9,8,4,0,1,7,null,null,null,2,5] (0's right child is 2 and 1's left child is 5)

     3
    /\
   /  \
   9   8
  /\  /\
 /  \/  \
 4  01   7
    /\
   /  \
   5   2

Output:

[
  [4],
  [9,5],
  [3,0,1],
  [8,2],
  [7]
]

思路：level order traversal (BFS) + hash table，ref: https://www.youtube.com/watch?v=PQKkr036wRc

定義horizontal distance: 從root出發，Hd(root) = 0, Hd(左孩子) = Hd(parent) - 1, Hd(右孩子) = Hd(parent) + 1

需要兩個hash map，dist記錄每個節點的hd, map記錄每個dist對應的節點的值。先把root放入佇列，在dist, map中更新root的相關資訊。當佇列不為空時，取出佇列首元素，check其左右孩子是否存在並更新子節點的hd（更新dist 和 map）。同時還需要兩個常數min, max記錄hd的範圍，以便最後能按順序把map中的list輸出

time: O(n), space: O(n)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    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<>();
        Map<TreeNode, Integer> dist = new HashMap<>();
        Queue<TreeNode> q = new LinkedList<>();
        int min = 0, max = 0;
        
        q.offer(root);
        dist.put(root, 0);
        map.put(0, new ArrayList<>());
        map.get(0).add(root.val);
        
        while(!q.isEmpty()) {
            TreeNode cur = q.poll();
            int dis = dist.get(cur);
            if(cur.left != null) {
                dist.put(cur.left, dis - 1);
                map.putIfAbsent(dis - 1, new ArrayList<>());
                map.get(dis - 1).add(cur.left.val);
                min = Math.min(min, dis - 1);
                q.offer(cur.left);
            }
            if(cur.right != null) {
                dist.put(cur.right, dis + 1);
                map.putIfAbsent(dis + 1, new ArrayList<>());
                map.get(dis + 1).add(cur.right.val);
                max = Math.max(max, dis + 1);
                q.offer(cur.right);
            }
        }

        for(int i = min; i <= max; i++) {
            res.add(map.get(i));
        }
        return res;
    }
}