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Leetcode314-binaryTreeVerticalOrderTraversal

Posted on Edited on In Views: Valine:
Solution Report of LeetCode Acceptted

Description

Given a binary tree, return the vertical order traversal of its nodes’ values. (ie, from top to bottom, column by column).

If two nodes are in the same row and column, the order should be from left to right.

Example

Examples 1:

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Input: [3,9,20,null,null,15,7]

   3
  /\
 /  \
 9  20
    /\
   /  \
  15   7 

Output:

[
  [9],
  [3,15],
  [20],
  [7]
]

Examples 2:
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Input: [3,9,8,4,0,1,7]

     3
    /\
   /  \
   9   8
  /\  /\
 /  \/  \
 4  01   7 

Output:

[
  [4],
  [9],
  [3,0,1],
  [8],
  [7]
]

Examples 3:
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Input: [3,9,8,4,0,1,7,null,null,null,2,5] (0's right child is 2 and 1's left child is 5)

     3
    /\
   /  \
   9   8
  /\  /\
 /  \/  \
 4  01   7
    /\
   /  \
   5   2

Output:

[
  [4],
  [9,5],
  [3,0,1],
  [8,2],
  [7]
]

Solution

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
public List<List<Integer>> verticalOrder(TreeNode root) {
    List<List<Integer>> res = new ArrayList<>();
    if (root == null) {
        return res;
    }
    
    Map<Integer, ArrayList<Integer>> map = new HashMap<>();
    Queue<TreeNode> q = new LinkedList<>();
    Queue<Integer> cols = new LinkedList<>();

    q.add(root); 
    cols.add(0);

    int min = 0;
    int max = 0;
    
    while (!q.isEmpty()) {
        TreeNode node = q.poll();
        int col = cols.poll();
        
        if (!map.containsKey(col)) {
            map.put(col, new ArrayList<Integer>());
        }
        map.get(col).add(node.val);

        if (node.left != null) {
            q.add(node.left); 
            cols.add(col - 1);
            min = Math.min(min, col - 1);
        }
        
        if (node.right != null) {
            q.add(node.right);
            cols.add(col + 1);
            max = Math.max(max, col + 1);
        }
    }

    for (int i = min; i <= max; i++) {
        res.add(map.get(i));
    }

    return res;
}