LeetCode 987: Vertical Order Traversal of a Binary Tree Solution

Master LeetCode problem 987 (Vertical Order Traversal of a Binary Tree), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

987. Vertical Order Traversal of a Binary Tree

Hard

Hash Table Tree Depth-First Search Breadth-First Search Sorting Binary Tree

Problem Explanation

Explanation:

To solve this problem, we can perform a level-order traversal of the binary tree while keeping track of the column indices for each node. We can use a TreeMap to store the nodes based on their column indices. For each node, we update its column index and row index based on its parent's indices.

Perform a level-order traversal of the binary tree using a queue.
Maintain a TreeMap to store nodes based on their column indices.
Update the column index and row index of each node based on its parent's indices.
Sort the nodes within each column based on their values.
Retrieve the vertical order traversal from the TreeMap.

Time Complexity: O(N * log(N)), where N is the number of nodes in the binary tree. Space Complexity: O(N).

Solution Code

import java.util.*;

class Solution {
    public List<List<Integer>> verticalTraversal(TreeNode root) {
        List<List<Integer>> result = new ArrayList<>();
        if (root == null) return result;

        TreeMap<Integer, TreeMap<Integer, List<Integer>>> map = new TreeMap<>();
        Queue<TreeNode> queue = new LinkedList<>();
        Queue<Integer> cols = new LinkedList<>();
        
        queue.offer(root);
        cols.offer(0);

        while (!queue.isEmpty()) {
            int size = queue.size();
            Map<Integer, List<Integer>> columnNodes = new HashMap<>();
            
            for (int i = 0; i < size; i++) {
                TreeNode node = queue.poll();
                int col = cols.poll();
                
                columnNodes.putIfAbsent(col, new ArrayList<>());
                columnNodes.get(col).add(node.val);
                
                if (node.left != null) {
                    queue.offer(node.left);
                    cols.offer(col - 1);
                }
                
                if (node.right != null) {
                    queue.offer(node.right);
                    cols.offer(col + 1);
                }
            }
            
            for (int key : columnNodes.keySet()) {
                map.putIfAbsent(key, new TreeMap<>());
                for (int val : columnNodes.get(key)) {
                    map.get(key).computeIfAbsent(size - 1, k -> new ArrayList<>()).add(val);
                }
            }
        }

        for (TreeMap<Integer, List<Integer>> colMap : map.values()) {
            List<Integer> column = new ArrayList<>();
            for (List<Integer> values : colMap.values()) {
                Collections.sort(values);
                column.addAll(values);
            }
            result.add(column);
        }

        return result;
    }
}

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