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]
]

Approach 1. recursion:

/**
 * Definition for a binary tree node. public class TreeNode { int val; TreeNode
 * left; TreeNode right; TreeNode(int x) { val = x; } }
 */
class Solution {
 public List> verticalOrder(TreeNode root) {
  TreeMap> map = new TreeMap>();
  int columnNum = 0;
  List> columnlist = new LinkedList>();
  inorder(root, map, columnNum);

  if (map != null) {
   Iterator ir = map.descendingKeySet().iterator();
   while (ir.hasNext()) {
    int key = (Integer) ir.next();
    columnlist.add(map.get(key));
   }
  }
  return columnlist;
 }

 private void inorder(TreeNode node, TreeMap> map, int columnNum) {
  if (node == null) {
   return;
  } else {
   List list = map.get(columnNum);
   if (list != null) {
    list.add(node.val);
   } else {
    list = new ArrayList();
    list.add(node.val);
    map.put(columnNum, list);
   }
  }
  if (node.left != null) {
   inorder(node.left, map, columnNum + 1);
  }

  if (node.right != null) {
   inorder(node.right, map, columnNum - 1);
  }
 }
}

Another approach:2

note: poll() : This method retrieves and removes the head (first element) of this list.
public List> verticalOrder(TreeNode root) {
    List> res = new ArrayList<>();
    if (root == null) {
        return res;
    }
    
    Map> map = new HashMap<>();
    Queue q = new LinkedList<>();
    Queue cols = new LinkedList<>();

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

    int min = 0;
    int max = 0;
    
    while (!q.isEmpty()) {
        TreeNode node = q.poll();
        int col = cols.poll();
        
        if (!map.containsKey(col)) {
            map.put(col, new ArrayList());
        }
        map.get(col).add(node.val);

        if (node.left != null) {
            q.add(node.left); 
            cols.add(col - 1);
            min = Math.min(min, col - 1);
        }
        
        if (node.right != null) {
            q.add(node.right);
            cols.add(col + 1);
            max = Math.max(max, col + 1);
        }
    }

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

    return res;
}