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.

Solution:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    class Node implements Comparable<Node> {
        int d;
        int v;
        
        public Node(int d, int v) {
            this.d = d;
            this.v = v;
        }
        
        @Override
        public int compareTo(Node n) {
            if (Integer.compare(d, n.d) == 0) {
                return Integer.compare(v, n.v);
            }
            return Integer.compare(d, n.d);
        }
    }
    
    public List<List<Integer>> verticalTraversal(TreeNode root) {
        Map<Integer, PriorityQueue<Node>> map = new HashMap();
        pre(root, map, 0, 0);
        List<List<Integer>> result = new ArrayList();
        for (int p = -500; p <= 500; p ++) {
            if (map.containsKey(p)) {
                PriorityQueue<Node> nodes = map.get(p);
                List<Integer> curr = new ArrayList();
                while (!nodes.isEmpty()) {
                    curr.add(nodes.poll().v);
                }
                result.add(curr);
            }
        }
        return result;
    }
    
    private void pre(TreeNode root, Map<Integer, PriorityQueue<Node>> map, int p, int d) {
        map.putIfAbsent(p, new PriorityQueue());
        map.get(p).add(new Node(d, root.val));
        if (root.left != null) {
            pre(root.left, map, p - 1, d + 1);
        }
        if (root.right != null) {
            pre(root.right, map, p + 1, d + 1);
        }
    }
}

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    Map<Integer, PriorityQueue<int[]>> map = new TreeMap();
    
    public List<List<Integer>> verticalTraversal(TreeNode root) {
        pre(root, 0, 0);   
        List<List<Integer>> result = new ArrayList();
        for (int h : map.keySet()) {
            List<Integer> list = new ArrayList();
            while (!map.get(h).isEmpty()) {
                list.add(map.get(h).poll()[1]);
            }
            result.add(list);
        }
        return result;
    }
    
    private void pre(TreeNode root, int d, int h) {
        if (root == null) return;
                                          // [d, val]
        map.putIfAbsent(h, new PriorityQueue<int[]>((a, b) -> a[0] - b[0] == 0 ? a[1] - b[1] : a[0] - b[0]));
        map.get(h).offer(new int[]{d, root.val});
        pre(root.left, d + 1, h - 1);
        pre(root.right, d + 1, h + 1);
    }
}