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.

/**
 * 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 x, y;
        public Node(TreeNode root, int x, int y) {
            this.root = root;
            this.x = x;
            this.y = y;
        }
    }
    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();
        Node c = new Node(root, 0, 0);
        Queue<Node> q = new LinkedList();
        q.offer(c);
        
        int minx = 0;
        int maxx = 0;
        while(!q.isEmpty()) {
            Node cur = q.poll();
            minx = Math.min(cur.x, minx);
            maxx = Math.max(cur.x, maxx);
            map.computeIfAbsent(cur.x, a -> new ArrayList()).add(cur);
            
            if(cur.root.left != null) {
                q.offer(new Node(cur.root.left, cur.x - 1, cur.y - 1));
            }
            if(cur.root.right != null) {
                q.offer(new Node(cur.root.right, cur.x + 1, cur.y - 1));
            }
        }
        
        int index = 0;
        for(int i = minx; i <= maxx; i++) {
            Collections.sort(map.get(i), (a, b) -> a.y == b.y ? a.root.val - b.root.val : b.y - a.y);
            
            res.add(new ArrayList());
            for(Node node : map.get(i)) {
                res.get(index).add(node.root.val);
            }
            index++;
        }
        return res;
    }
}

https://www.youtube.com/watch?v=AUQL0R-ibEw

首先重新弄一个Node，包含了TreeNode和x、y坐标

然后用bfs来遍历所有的点并添加到map（x coord，List）中。因为我们最后要求从x最小到最大输出，所以为方便起见直接把minx和maxx持续更新了

因为可能两个点的位置相同，所以我们按x输出时，比较两个点的y，如果相同就先输出val小的那个，否则先输出高度高的。