314. Binary Tree Vertical Order Traversal

Given a binary tree, return thevertical ordertraversal 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]
]

FB: Complexity!

Thoughts

1. Use map <col, list<val>> to record the col and list value pair. Use BFS to expand the tree to add entry

Code T O(V), S O(V)

/**
 * 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, ArrayList<Integer>> map = new HashMap();
        Queue<TreeNode> q = new LinkedList<>();
        Queue<Integer> cols = new LinkedList<>();

        int min = 0, max = 0;

        q.add(root); cols.add(0);

        while(!q.isEmpty()){
            TreeNode cur = q.poll();
            int j = cols.poll();

            // add <col, val> pair            
            if(!map.containsKey(j)) map.put(j, new ArrayList<Integer>());
            map.get(j).add(cur.val);

            // update the range of col value
            min = Math.min(min, j);
            max = Math.max(max, j);

            // expand children
            if(cur.left != null){
                q.add(cur.left);
                cols.add(j - 1);
            }

            if(cur.right != null){
                q.add(cur.right);
                cols.add(j+ 1);
            }

        }

        // add all the list entry
        for (int i = min ; i <= max; i++ ){
            res.add(map.get(i));
        }

        return res;
    }
}

Python

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def verticalOrder(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        cols = collections.defaultdict(list)
        queue = [(root,0)]

        for node, col in queue:
            if node:
                cols[col].append(node.val)
                queue += (node.left, col - 1), (node.right, col + 1)

        return [cols[col] for col in sorted(cols)]