_987.java

package com.fishercoder.solutions;

import com.fishercoder.common.classes.TreeNode;

import java.util.List;
import java.util.ArrayList;
import java.util.PriorityQueue;
import java.util.TreeMap;

/**
 * 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:
 *
 *        3
 *       / \
 *      9  20
 *        /  \
 *       15  7
 *
 * 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:
 *
 *       1
 *      / \
 *     2   3
 *    / \ / \
 *   4  5 6 7
 *
 * 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.
 * */
public class _987 {
    public static class Solution1 {
        /**credit: https://leetcode.com/problems/vertical-order-traversal-of-a-binary-tree/discuss/231148/Java-TreeMap-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);
        }
    }
}