987.Vertical-Order-Traversal-of-a-Binary-Tree

987. Vertical Order Traversal of a Binary Tree

题目地址

https://leetcode.com/problems/vertical-order-traversal-of-a-binary-tree/

题目描述

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:
The tree will have between 1 and 1000 nodes.
Each node's value will be between 0 and 1000.

代码

Approach #2 Nested TreeMap + PriorityQueue => (x, y, root.val)

class Solution {
    public List<List<Integer>> verticalTraversal(TreeNode root) {
        TreeMap<Integer, TreeMap<Integer, PriorityQueue<Integer>>> map = new TreeMap<>();
        dfs(root, 0, 0, map);
        List<List<Integer>> list = new ArrayList<>();
        for (TreeMap<Integer, PriorityQueue<Integer>> ys : map.values()) {
            list.add(new ArrayList<>());
            for (PriorityQueue<Integer> nodes : ys.values()) {
                while (!nodes.isEmpty()) {
                    list.get(list.size() - 1).add(nodes.poll());
                }
            }
        }
        return list;
    }
    private void dfs(TreeNode root, int x, int y, TreeMap<Integer, TreeMap<Integer, PriorityQueue<Integer>>> map) {
        if (root == null)         return;

        if (!map.containsKey(x)) {
            map.put(x, new TreeMap<>());
        }
        if (!map.get(x).containsKey(y)) {
            map.get(x).put(y, new PriorityQueue<>());
        }
        map.get(x).get(y).offer(root.val);
        dfs(root.left, x - 1, y + 1, map);
        dfs(root.right, x + 1, y + 1, map);
    }
}

Approach #1 Store Locations

class Solution {
    List<Location> locations;
    public List<List<Integer>> verticalTraversal(TreeNode root) {
        // Each location is a node's x position, y position, and value
        locations = new ArrayList();
        dfs(root, 0, 0);
        Collections.sort(locations);

        List<List<Integer>> ans = new ArrayList();
        ans.add(new ArrayList<Integer>());

        int prev = locations.get(0).x;

        for (Location loc: locations) {
            // If the x value changed, it's part of a new report.
            if (loc.x != prev) {
                prev = loc.x;
                ans.add(new ArrayList<Integer>());
            }

            // We always add the node's value to the latest report.
            ans.get(ans.size() - 1).add(loc.val);
        }

        return ans;
    }

    public void dfs(TreeNode node, int x, int y) {
        if (node != null) {
            locations.add(new Location(x, y, node.val));
            dfs(node.left, x-1, y+1);
            dfs(node.right, x+1, y+1);
        }
    }
}

class Location implements Comparable<Location>{
    int x, y, val;
    Location(int x, int y, int val) {
        this.x = x;
        this.y = y;
        this.val = val;
    }

    @Override
    public int compareTo(Location that) {
        if (this.x != that.x)
            return Integer.compare(this.x, that.x);
        else if (this.y != that.y)
            return Integer.compare(this.y, that.y);
        else
            return Integer.compare(this.val, that.val);
    }
}