987. 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.

 

和 314. Binary Tree Vertical Order Traversal 的区别:

314. If two nodes are in the same row and column, the order should be from left to right.

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

time = O(n), space = O(n)

/**
 * 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 {
        TreeNode root;
        int hd, vd;
        public Node(TreeNode root, int hd, int vd) {
            this.root = root;
            this.hd = hd;
            this.vd = vd;
        }
    }
    public List<List<Integer>> verticalTraversal(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        if(root == null) {
            return res;
        }
        
        Map<Integer, List<Node>> map = new HashMap<>();
        Queue<Node> q = new LinkedList<>();
        int minHd = 0, maxHd = 0;
        
        q.offer(new Node(root, 0, 0));
        
        while(!q.isEmpty()) {
            Node cur = q.poll();
            minHd = Math.min(minHd, cur.hd);
            maxHd = Math.max(maxHd, cur.hd);
            
            map.putIfAbsent(cur.hd, new ArrayList<>());
            map.get(cur.hd).add(cur);
            
            if(cur.root.left != null) {
                q.offer(new Node(cur.root.left, cur.hd - 1, cur.vd + 1));
            }
            if(cur.root.right != null) {
                q.offer(new Node(cur.root.right, cur.hd + 1, cur.vd + 1));
            }
        }
        
        int idx = 0;
        for(int i = minHd; i <= maxHd; i++) {
            Collections.sort(map.get(i), (a, b) -> {
                if(a.vd - b.vd == 0) {
                    return a.root.val - b.root.val;
                } else {
                    return a.vd - b.vd;
                }
            });
            res.add(new ArrayList<>());
            for(Node node : map.get(i)) {
                res.get(idx).add(node.root.val);
            }
            idx++;
        }
        return res;
    }
}