# Vertical Order Traversal of a Binary Tree – Leetcode Solution

In this post, we are going to solve the 987. Vertical Order Traversal of a Binary Tree problem of Leetcode. This problem 987. Vertical Order Traversal of a Binary Tree is a Leetcode hard level problem.

Let’s see the code, 987. Vertical Order Traversal of a Binary Tree – Leetcode Solution.

## Problem

Given the `root` of a binary tree, calculate the vertical order traversal of the binary tree.

For each node at position `(row, col)`, its left and right children will be at positions `(row + 1, col - 1)` and `(row + 1, col + 1) `respectively. The root of the tree is at `(0, 0)`.

The vertical order traversal of a binary tree is a list of top-to-bottom orderings for each column index starting from the leftmost column and ending on the rightmost column. There may be multiple nodes in the same row and same column. In such a case, sort these nodes by their values.

Return the vertical order traversal of the binary tree.

### Example 1 :

``````Input: root = [3,9,20,null,null,15,7]
Output: [[9],[3,15],[20],[7]]
Explanation:
Column -1: Only node 9 is in this column.
Column 0: Nodes 3 and 15 are in this column in that order from top to bottom.
Column 1: Only node 20 is in this column.
Column 2: Only node 7 is in this column.``````

### Example 2 :

``````Input: root = [1,2,3,4,5,6,7]
Output: [[4],[2],[1,5,6],[3],[7]]
Explanation:
Column -2: Only node 4 is in this column.
Column -1: Only node 2 is in this column.
Column 0: Nodes 1, 5, and 6 are in this column.
1 is at the top, so it comes first.
5 and 6 are at the same position (2, 0), so we order them by their value, 5 before 6.
Column 1: Only node 3 is in this column.
Column 2: Only node 7 is in this column.``````

### Example 3 :

``````Input: root = [1,2,3,4,6,5,7]
Output: [[4],[2],[1,5,6],[3],[7]]
Explanation:
This case is the exact same as example 2, but with nodes 5 and 6 swapped.
Note that the solution remains the same since 5 and 6 are in the same location and should be ordered by their values.``````

### Constraints

• The number of nodes in the tree is in the range `[1, 1000]`.
• `0 <= Node.val <= 1000.`

Now, let’s see the code of 987. Vertical Order Traversal of a Binary Tree – Leetcode Solution.

# Vertical Order Traversal of a Binary Tree – Leetcode Solution

### 987. Vertical Order Traversal of a Binary Tree – Solution in Java

```/**
* 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) {
TreeMap<Integer, TreeMap<Integer, PriorityQueue<Integer>>> map = new TreeMap<>();
dfs(root, 0, 0, map);
List<List<Integer>> list = new ArrayList<>();
for (TreeMap<Integer, PriorityQueue<Integer>> ys : map.values()) {
for (PriorityQueue<Integer> nodes : ys.values()) {
while (!nodes.isEmpty()) {
}
}
}
return list;
}
private void dfs(TreeNode root, int x, int y, TreeMap<Integer, TreeMap<Integer, PriorityQueue<Integer>>> map) {
if (root == null) {
return;
}
if (!map.containsKey(x)) {
map.put(x, new TreeMap<>());
}
if (!map.get(x).containsKey(y)) {
map.get(x).put(y, new PriorityQueue<>());
}
map.get(x).get(y).offer(root.val);
dfs(root.left, x - 1, y + 1, map);
dfs(root.right, x + 1, y + 1, map);
}
}
```

### 987. Vertical Order Traversal of a Binary Tree – Solution in C++

```/**
* 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:
vector<vector<int>> verticalTraversal(TreeNode* root) {
map<int, map<int, set<int>>> nodes;
traverse(root, 0, 0, nodes);
vector<vector<int>> ans;
for (auto p : nodes) {
vector<int> col;
for (auto q : p.second) {
col.insert(col.end(), q.second.begin(), q.second.end());
}
ans.push_back(col);
}
return ans;
}
private:
void traverse(TreeNode* root, int x, int y, map<int, map<int, set<int>>>& nodes) {
if (root) {
nodes[x][y].insert(root -> val);
traverse(root -> left, x - 1, y + 1, nodes);
traverse(root -> right, x + 1, y + 1, nodes);
}
}
};```

### 987. Vertical Order Traversal of a Binary Tree– Solution in Python

```# 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
class Solution(object):
def verticalTraversal(self, root):
g = collections.defaultdict(list)
queue = [(root,0)]
while queue:
new = []
d = collections.defaultdict(list)
for node, s in queue:
d[s].append(node.val)
if node.left:  new += (node.left, s-1),
if node.right: new += (node.right,s+1),
for i in d: g[i].extend(sorted(d[i]))
queue = new
return [g[i] for i in sorted(g)]
```

Note: This problem 987. Vertical Order Traversal of a Binary Tree is generated by Leetcode but the solution is provided by CodingBroz. This tutorial is only for Educational and Learning purpose.