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| 1 | +package com.fishercoder.solutions; |
| 2 | + |
| 3 | +import com.fishercoder.common.classes.TreeNode; |
| 4 | + |
| 5 | +import java.util.*; |
| 6 | + |
| 7 | +/** |
| 8 | + * 987. Vertical Order Traversal of a Binary Tree |
| 9 | + * |
| 10 | + * Given a binary tree, return the vertical order traversal of its nodes values. |
| 11 | + * 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). |
| 12 | + * 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). |
| 13 | + * If two nodes have the same position, then the value of the node that is reported first is the value that is smaller. |
| 14 | + * Return an list of non-empty reports in order of X coordinate. Every report will have a list of values of nodes. |
| 15 | + * |
| 16 | + * Example 1: |
| 17 | + * |
| 18 | + * 3 |
| 19 | + * / \ |
| 20 | + * 9 20 |
| 21 | + * / \ |
| 22 | + * 15 7 |
| 23 | + * |
| 24 | + * Input: [3,9,20,null,null,15,7] |
| 25 | + * Output: [[9],[3,15],[20],[7]] |
| 26 | + * Explanation: |
| 27 | + * Without loss of generality, we can assume the root node is at position (0, 0): |
| 28 | + * Then, the node with value 9 occurs at position (-1, -1); |
| 29 | + * The nodes with values 3 and 15 occur at positions (0, 0) and (0, -2); |
| 30 | + * The node with value 20 occurs at position (1, -1); |
| 31 | + * The node with value 7 occurs at position (2, -2). |
| 32 | + * |
| 33 | + * |
| 34 | + * Example 2: |
| 35 | + * |
| 36 | + * 1 |
| 37 | + * / \ |
| 38 | + * 2 3 |
| 39 | + * / \ / \ |
| 40 | + * 4 5 6 7 |
| 41 | + * |
| 42 | + * Input: [1,2,3,4,5,6,7] |
| 43 | + * Output: [[4],[2],[1,5,6],[3],[7]] |
| 44 | + * Explanation: |
| 45 | + * The node with value 5 and the node with value 6 have the same position according to the given scheme. |
| 46 | + * However, in the report "[1,5,6]", the node value of 5 comes first since 5 is smaller than 6. |
| 47 | + * |
| 48 | + * Note: |
| 49 | + * |
| 50 | + * The tree will have between 1 and 1000 nodes. |
| 51 | + * Each node's value will be between 0 and 1000. |
| 52 | + * */ |
| 53 | +public class _987 { |
| 54 | + public static class Solution1 { |
| 55 | + /**credit: https://leetcode.com/problems/vertical-order-traversal-of-a-binary-tree/discuss/231148/Java-TreeMap-Solution*/ |
| 56 | + public List<List<Integer>> verticalTraversal(TreeNode root) { |
| 57 | + TreeMap<Integer, TreeMap<Integer, PriorityQueue<Integer>>> map = new TreeMap<>(); |
| 58 | + dfs(root, 0, 0, map); |
| 59 | + List<List<Integer>> list = new ArrayList<>(); |
| 60 | + for (TreeMap<Integer, PriorityQueue<Integer>> ys : map.values()) { |
| 61 | + list.add(new ArrayList<>()); |
| 62 | + for (PriorityQueue<Integer> nodes : ys.values()) { |
| 63 | + while (!nodes.isEmpty()) { |
| 64 | + list.get(list.size() - 1).add(nodes.poll()); |
| 65 | + } |
| 66 | + } |
| 67 | + } |
| 68 | + return list; |
| 69 | + } |
| 70 | + |
| 71 | + private void dfs(TreeNode root, int x, int y, TreeMap<Integer, TreeMap<Integer, PriorityQueue<Integer>>> map) { |
| 72 | + if (root == null) { |
| 73 | + return; |
| 74 | + } |
| 75 | + if (!map.containsKey(x)) { |
| 76 | + map.put(x, new TreeMap<>()); |
| 77 | + } |
| 78 | + if (!map.get(x).containsKey(y)) { |
| 79 | + map.get(x).put(y, new PriorityQueue<>()); |
| 80 | + } |
| 81 | + map.get(x).get(y).offer(root.val); |
| 82 | + dfs(root.left, x - 1, y + 1, map); |
| 83 | + dfs(root.right, x + 1, y + 1, map); |
| 84 | + } |
| 85 | + } |
| 86 | +} |
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