314. Binary Tree Vertical Order Traversal

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

定义horizontal distance: 从root出发，Hd(root) = 0, Hd(左孩子) = Hd(parent) - 1, Hd(右孩子) = Hd(parent) + 1

需要两个hash map，dist记录每个节点的hd, map记录每个dist对应的节点的值。先把root放入队列，在dist, map中更新root的相关信息。当队列不为空时，取出队列首元素，check其左右孩子是否存在并更新子节点的hd（更新dist 和 map）。同时还需要两个常数min, max记录hd的范围，以便最后能按顺序把map中的list输出

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;
    }
}