📚 Problem Overview

In Vertical Order Traversal of a Binary Tree, we need to:

Print nodes column-wise from left to right.
Sort nodes in each column top to bottom.
If two nodes share the same position, sort their values in ascending order.

💡 Thought Process

We can imagine the binary tree as plotted on a 2D grid:

The root node starts at (0, 0).
Left child: moves to (verticalIndex - 1, level + 1)
Right child: moves to (verticalIndex + 1, level + 1)

We must:

Group nodes by vertical index (columns).
Sort nodes by level (rows).
Sort overlapping nodes (same vertical and level) by value.

✨ Example

Let’s consider this tree:

Vertical Index Mapping:

Vertical Index	Nodes
-1	9
0	3, 15
1	20
2	7

Final Vertical Order Traversal:

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

Why?

Vertical -1: Only 9.
Vertical 0: 3 is on top, 15 is below → [3, 15].
Vertical 1: Only 20.
Vertical 2: Only 7.

🚀 Approach

We’ll use:

TreeMap<verticalIndex, TreeMap<level, PriorityQueue<node values>>>
→ This helps automatically sort verticals, levels, and node values.
DFS traversal to map each node to its (verticalIndex, level).

✅ Java Solution

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<List<Integer>> verticalTraversal(TreeNode root) {
        List<List<Integer>> result = new ArrayList<>();
        if (root == null) return result;

        TreeMap<Integer, TreeMap<Integer, PriorityQueue<Integer>>> map = new TreeMap<>();
        traverse(root, map, 0, 0);

        for (TreeMap<Integer, PriorityQueue<Integer>> levels : map.values()) {
            List<Integer> col = new ArrayList<>();
            for (PriorityQueue<Integer> nodes : levels.values()) {
                while (!nodes.isEmpty()) {
                    col.add(nodes.poll());
                }
            }
            result.add(col);
        }
        return result;
    }

    private void traverse(TreeNode root, TreeMap<Integer, TreeMap<Integer, PriorityQueue<Integer>>> map, int level, int verticalIndex) {
        if (root == null) return;

        map.putIfAbsent(verticalIndex, new TreeMap<>());
        map.get(verticalIndex).putIfAbsent(level, new PriorityQueue<>());

        map.get(verticalIndex).get(level).offer(root.val);

        traverse(root.left, map, level + 1, verticalIndex - 1);
        traverse(root.right, map, level + 1, verticalIndex + 1);
    }
}

✅ C++ Solution

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    map<int, map<int, multiset<int>>> nodes;

    void dfs(TreeNode* node, int verticalIndex, int level) {
        if (!node) return;

        nodes[verticalIndex][level].insert(node->val);

        dfs(node->left, verticalIndex - 1, level + 1);
        dfs(node->right, verticalIndex + 1, level + 1);
    }

    vector<vector<int>> verticalTraversal(TreeNode* root) {
        vector<vector<int>> result;
        dfs(root, 0, 0);

        for (auto& p : nodes) {
            vector<int> col;
            for (auto& q : p.second) {
                col.insert(col.end(), q.second.begin(), q.second.end());
            }
            result.push_back(col);
        }
        return result;
    }
};

✅ Python Solution

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right

from collections import defaultdict
import heapq

class Solution:
    def verticalTraversal(self, root):
        nodes = defaultdict(lambda: defaultdict(list))

        def dfs(node, verticalIndex, level):
            if not node:
                return
            heapq.heappush(nodes[verticalIndex][level], node.val)
            dfs(node.left, verticalIndex - 1, level + 1)
            dfs(node.right, verticalIndex + 1, level + 1)

        dfs(root, 0, 0)

        result = []
        for verticalIndex in sorted(nodes.keys()):
            col = []
            for level in sorted(nodes[verticalIndex].keys()):
                while nodes[verticalIndex][level]:
                    col.append(heapq.heappop(nodes[verticalIndex][level]))
            result.append(col)

        return result

🔍 Key Notes

We use TreeMap (Java), map/multiset (C++), and dict/heapq (Python) to maintain sorting.
DFS traversal is essential to track vertical and level positions.

🕐 Complexity

Time Complexity: O(N log N)
- Sorting verticals, levels, and node values.
Space Complexity: O(N)
- Additional space for maps and priority queues.

🙌 Final Thoughts

Vertical Order Traversal is a great problem to strengthen:

Binary Tree Traversal
Coordinate Mapping in Grids
Efficient Sorting using Maps and Priority Queues

If you’re aiming to master tree problems, this one will give you a solid foundation!

📢 Tags:

#LeetCode #Java #C++ #Python #DSA #BinaryTree #VerticalOrderTraversal #100DaysOfCode #Algorithms

🌳 Vertical Order Traversal of Binary Tree – Complete Guide with Java, C++, and Python Solutions