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. Given binary tree [3,9,20,null,null,15,7],
   

      3
     /
    /  
    9  20
       /
      /  
     15   7
   

   return its vertical order traversal as:

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

2. Given binary tree [3,9,8,4,0,1,7],
   

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

   return its vertical order traversal as:

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

3. Given binary tree [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
   

   return its vertical order traversal as:

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

Analyses: Do BFS, have the root column as 0, record the leftmost column and rightmost column. Map the column as the node values in that column. 

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<List<Integer>> verticalOrder(TreeNode root) {
        List<List<Integer> > result = new ArrayList<>();
        if (root == null) return result;
        
        // map the column to the nodes values in that column
        HashMap<Integer, List<Integer> > map = new HashMap<Integer, List<Integer> >();
        Queue<TreeNode> nodes = new LinkedList<TreeNode>();
        Queue<Integer> columns = new LinkedList<Integer>();
        
        nodes.add(root);
        columns.add(0);
        int left = 0, right = 0;
        while (nodes.peek() != null) {
            TreeNode node = nodes.remove();
            int column = columns.remove();
            
            if (!map.containsKey(column)) map.put(column, new ArrayList<Integer>());
            map.get(column).add(node.val);
            
            if (node.left != null) {
                nodes.add(node.left);
                columns.add(column - 1);
                left = Math.min(left, column - 1);
            }
            
            if (node.right != null) {
                nodes.add(node.right);
                columns.add(column + 1);
                right = Math.max(right, column + 1);
            }
        }
        
        for (int i = left; i <= right; i++) {
            List<Integer> temp = map.get(i);
            result.add(temp);
        }
        return result;
    }
}