987. Vertical Order Traversal of a Binary Tree.md

987. Vertical Order Traversal of a Binary Tree

Question:

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.

Solution:

• Method 1: DFS + TreeMap, dfs is for giving coordinators to nodes and TreeMap is for order. *Time complexity: dfs(O(Vertice + Edge)) + TreeMap(O(NlogN))

  /**
   * Definition for a binary tree node.
   * public class TreeNode {
   *     int val;
   *     TreeNode left;
   *     TreeNode right;
   *     TreeNode(int x) { val = x; }
   * }
   */
  class Solution {
      private TreeMap<Integer, TreeMap<Integer, PriorityQueue<Integer>>> map;
      public List<List<Integer>> verticalTraversal(TreeNode root) {
          map = new TreeMap<>();
          List<List<Integer>> result = new ArrayList<>();
          if(root == null) return result;
          dfs(0, 0, root);
          for(TreeMap<Integer, PriorityQueue<Integer>> value : map.values()){
              List<Integer> temp = new ArrayList<>();
              for(PriorityQueue<Integer> pq : value.values()){
                  while(!pq.isEmpty()){
                      temp.add(pq.poll());
                  }
              }
              result.add(temp);
          }
          return result;
      }
      private void dfs(int x, int y, TreeNode node){
          TreeMap<Integer, PriorityQueue<Integer>> value = map.containsKey(x) ? map.get(x): new TreeMap<Integer, PriorityQueue<Integer>>(
              new Comparator<Integer>(){
                  @Override
                  public int compare(Integer n1, Integer n2){
                      return n2 - n1;
                  }
              }
          );
          PriorityQueue<Integer> pq = value.containsKey(y) ? value.get(y): new PriorityQueue<Integer>();
          pq.offer(node.val);
          value.put(y, pq);
          map.put(x, value);
          if(node.left != null) dfs(x - 1, y - 1, node.left);
          if(node.right != null) dfs(x + 1, y - 1, node.right);
      }
  }