Vertical Order Traversal of a Binary Tree

Given a binary tree, return the vertical order traversal of its nodes values.

For each node at position (X, Y), its left and right children respectively will be at positions (X-1, Y-1) and (X+1, Y-1).

Running a vertical line from X = -infinity to X = +infinity, whenever the vertical line touches some nodes, we report the values of the nodes in order from top to bottom (decreasing Y coordinates).

If two nodes have the same position, then the value of the node that is reported first is the value that is smaller.

Return an list of non-empty reports in order of X coordinate. Every report will have a list of values of nodes.

Example 1:

Input: [3,9,20,null,null,15,7]
Output: [[9],[3,15],[20],[7]]
Explanation: 
Without loss of generality, we can assume the root node is at position (0, 0):
Then, the node with value 9 occurs at position (-1, -1);
The nodes with values 3 and 15 occur at positions (0, 0) and (0, -2);
The node with value 20 occurs at position (1, -1);
The node with value 7 occurs at position (2, -2).

Example 2:

Input: [1,2,3,4,5,6,7]
Output: [[4],[2],[1,5,6],[3],[7]]
Explanation: 
The node with value 5 and the node with value 6 have the same position according to the given scheme.
However, in the report "[1,5,6]", the node value of 5 comes first since 5 is smaller than 6.

Note:

1. The tree will have between 1 and 1000 nodes.

2. Each node's value will be between 0 and 1000.

class Solution {
    static class Pair {
        TreeNode root;
        int level;

        Pair(TreeNode node, int level) {
            this.root = node;
            this.level = level;
        }
    }

    public List<List<Integer>> verticalTraversal(TreeNode root) {
        // Map for storing level -> List
        HashMap<Integer, List<Integer>> map = new HashMap<>();
        int minLevel = Integer.MAX_VALUE, maxLevel = Integer.MIN_VALUE;
        // Going for level order BFS
        Queue<Pair> q = new LinkedList<>();
        q.add(new Pair(root, 0));
        while (q.size() != 0) {
            int size = q.size();
            List<Pair> level = new ArrayList<>();
            while (size-- > 0) {
                Pair node = q.poll();
                minLevel = Math.min(minLevel, node.level);
                maxLevel = Math.max(maxLevel, node.level);
                map.computeIfAbsent(node.level, k -> new ArrayList<>()).add(node.root.val);
                if (node.root.left != null)
                    level.add(new Pair(node.root.left, node.level - 1));
                if (node.root.right != null)
                    level.add(new Pair(node.root.right, node.level + 1));
            }
            // Sorting levels on values
            // Because of this -> If two nodes have the same position, then the value of the
            // node that is reported first is the value that is smaller.
            Collections.sort(level, (a, b) -> a.root.val - b.root.val);
            for (Pair node : level)
                q.add(node);
        }
        List<List<Integer>> ans = new ArrayList<>();
        for (int i = minLevel; i <= maxLevel; i++)
            ans.add(map.get(i));
        return ans;
    }
}