*Binary Tree Vertical Order Traversal

 1 public class Solution {
 2     public List<List<Integer>> verticalOrder(TreeNode root) {
 3         List<List<Integer>> res = new ArrayList<>();
 4         if (root == null) {
 5             return res;
 6         }
 7        //map‘s key is column, we assume the root column is zero, the left node will minus 1 ,and the right node will plus 1
 8         HashMap<Integer, ArrayList<Integer>> map = new HashMap<Integer, ArrayList<Integer>>();
 9         Queue<TreeNode> queue = new LinkedList<>();
10        //use a HashMap to store the TreeNode and the according cloumn value
11         HashMap<TreeNode, Integer> weight = new HashMap<TreeNode, Integer>();
12         queue.offer(root);
13         weight.put(root, 0);
14         int min = 0;
15         while (!queue.isEmpty()) {
16             TreeNode node = queue.poll();
17             int w = weight.get(node);
18             if (!map.containsKey(w)) {
19                 map.put(w, new ArrayList<>());
20             }
21             map.get(w).add(node.val);
22             if (node.left != null) {
23                 queue.add(node.left);
24                 weight.put(node.left, w - 1);
25             } 
26             if (node.right != null) {
27                 queue.add(node.right);
28                 weight.put(node.right, w + 1);
29             }
30             //update min ,min means the minimum column value, which is the left most node
31             min = Math.min(min, w);
32         }
33         while (map.containsKey(min)) {
34             res.add(map.get(min++));
35         }
36         return res;
37     }
38 }