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            "title": "Vertical Order Traversal of a Binary Tree",
            "titleSlug": "vertical-order-traversal-of-a-binary-tree",
            "content": "<p>Given the <code>root</code> of a binary tree, calculate the <strong>vertical order traversal</strong> of the binary tree.</p>\n\n<p>For each node at position <code>(row, col)</code>, its left and right children will be at positions <code>(row + 1, col - 1)</code> and <code>(row + 1, col + 1)</code> respectively. The root of the tree is at <code>(0, 0)</code>.</p>\n\n<p>The <strong>vertical order traversal</strong> 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.</p>\n\n<p>Return <em>the <strong>vertical order traversal</strong> of the binary tree</em>.</p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/01/29/vtree1.jpg\" style=\"width: 431px; height: 304px;\" />\n<pre>\n<strong>Input:</strong> root = [3,9,20,null,null,15,7]\n<strong>Output:</strong> [[9],[3,15],[20],[7]]\n<strong>Explanation:</strong>\nColumn -1: Only node 9 is in this column.\nColumn 0: Nodes 3 and 15 are in this column in that order from top to bottom.\nColumn 1: Only node 20 is in this column.\nColumn 2: Only node 7 is in this column.</pre>\n\n<p><strong>Example 2:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/01/29/vtree2.jpg\" style=\"width: 512px; height: 304px;\" />\n<pre>\n<strong>Input:</strong> root = [1,2,3,4,5,6,7]\n<strong>Output:</strong> [[4],[2],[1,5,6],[3],[7]]\n<strong>Explanation:</strong>\nColumn -2: Only node 4 is in this column.\nColumn -1: Only node 2 is in this column.\nColumn 0: Nodes 1, 5, and 6 are in this column.\n          1 is at the top, so it comes first.\n          5 and 6 are at the same position (2, 0), so we order them by their value, 5 before 6.\nColumn 1: Only node 3 is in this column.\nColumn 2: Only node 7 is in this column.\n</pre>\n\n<p><strong>Example 3:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/01/29/vtree3.jpg\" style=\"width: 512px; height: 304px;\" />\n<pre>\n<strong>Input:</strong> root = [1,2,3,4,6,5,7]\n<strong>Output:</strong> [[4],[2],[1,5,6],[3],[7]]\n<strong>Explanation:</strong>\nThis case is the exact same as example 2, but with nodes 5 and 6 swapped.\nNote that the solution remains the same since 5 and 6 are in the same location and should be ordered by their values.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li>The number of nodes in the tree is in the range <code>[1, 1000]</code>.</li>\n\t<li><code>0 &lt;= Node.val &lt;= 1000</code></li>\n</ul>\n",
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                    "name": "Hash Table",
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                    "name": "Tree",
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                    "code": "/**\n * Definition for a binary tree node.\n * struct TreeNode {\n *     int val;\n *     TreeNode *left;\n *     TreeNode *right;\n *     TreeNode() : val(0), left(nullptr), right(nullptr) {}\n *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}\n *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}\n * };\n */\nclass Solution {\npublic:\n    vector<vector<int>> verticalTraversal(TreeNode* root) {\n        \n    }\n};",
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                    "code": "/**\n * Definition for a binary tree node.\n * public class TreeNode {\n *     int val;\n *     TreeNode left;\n *     TreeNode right;\n *     TreeNode() {}\n *     TreeNode(int val) { this.val = val; }\n *     TreeNode(int val, TreeNode left, TreeNode right) {\n *         this.val = val;\n *         this.left = left;\n *         this.right = right;\n *     }\n * }\n */\nclass Solution {\n    public List<List<Integer>> verticalTraversal(TreeNode root) {\n        \n    }\n}",
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                    "lang": "Python",
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                    "code": "# Definition for a binary tree node.\n# class TreeNode(object):\n#     def __init__(self, val=0, left=None, right=None):\n#         self.val = val\n#         self.left = left\n#         self.right = right\nclass Solution(object):\n    def verticalTraversal(self, root):\n        \"\"\"\n        :type root: TreeNode\n        :rtype: List[List[int]]\n        \"\"\"\n        ",
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                    "code": "# Definition for a binary tree node.\n# class TreeNode:\n#     def __init__(self, val=0, left=None, right=None):\n#         self.val = val\n#         self.left = left\n#         self.right = right\nclass Solution:\n    def verticalTraversal(self, root: Optional[TreeNode]) -> List[List[int]]:\n        ",
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                    "code": "/**\n * Definition for a binary tree node.\n * struct TreeNode {\n *     int val;\n *     struct TreeNode *left;\n *     struct TreeNode *right;\n * };\n */\n\n\n/**\n * Return an array of arrays of size *returnSize.\n * The sizes of the arrays are returned as *returnColumnSizes array.\n * Note: Both returned array and *columnSizes array must be malloced, assume caller calls free().\n */\nint** verticalTraversal(struct TreeNode* root, int* returnSize, int** returnColumnSizes){\n\n}",
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                    "code": "/**\n * Definition for a binary tree node.\n * public class TreeNode {\n *     public int val;\n *     public TreeNode left;\n *     public TreeNode right;\n *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {\n *         this.val = val;\n *         this.left = left;\n *         this.right = right;\n *     }\n * }\n */\npublic class Solution {\n    public IList<IList<int>> VerticalTraversal(TreeNode root) {\n        \n    }\n}",
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                    "code": "/**\n * Definition for a binary tree node.\n * function TreeNode(val, left, right) {\n *     this.val = (val===undefined ? 0 : val)\n *     this.left = (left===undefined ? null : left)\n *     this.right = (right===undefined ? null : right)\n * }\n */\n/**\n * @param {TreeNode} root\n * @return {number[][]}\n */\nvar verticalTraversal = function(root) {\n    \n};",
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                    "code": "# Definition for a binary tree node.\n# class TreeNode\n#     attr_accessor :val, :left, :right\n#     def initialize(val = 0, left = nil, right = nil)\n#         @val = val\n#         @left = left\n#         @right = right\n#     end\n# end\n# @param {TreeNode} root\n# @return {Integer[][]}\ndef vertical_traversal(root)\n    \nend",
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                    "code": "/**\n * Definition for a binary tree node.\n * public class TreeNode {\n *     public var val: Int\n *     public var left: TreeNode?\n *     public var right: TreeNode?\n *     public init() { self.val = 0; self.left = nil; self.right = nil; }\n *     public init(_ val: Int) { self.val = val; self.left = nil; self.right = nil; }\n *     public init(_ val: Int, _ left: TreeNode?, _ right: TreeNode?) {\n *         self.val = val\n *         self.left = left\n *         self.right = right\n *     }\n * }\n */\nclass Solution {\n    func verticalTraversal(_ root: TreeNode?) -> [[Int]] {\n        \n    }\n}",
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                    "code": "/**\n * Definition for a binary tree node.\n * type TreeNode struct {\n *     Val int\n *     Left *TreeNode\n *     Right *TreeNode\n * }\n */\nfunc verticalTraversal(root *TreeNode) [][]int {\n    \n}",
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                    "code": "/**\n * Example:\n * var ti = TreeNode(5)\n * var v = ti.`val`\n * Definition for a binary tree node.\n * class TreeNode(var `val`: Int) {\n *     var left: TreeNode? = null\n *     var right: TreeNode? = null\n * }\n */\nclass Solution {\n    fun verticalTraversal(root: TreeNode?): List<List<Int>> {\n        \n    }\n}",
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                    "code": "// Definition for a binary tree node.\n// #[derive(Debug, PartialEq, Eq)]\n// pub struct TreeNode {\n//   pub val: i32,\n//   pub left: Option<Rc<RefCell<TreeNode>>>,\n//   pub right: Option<Rc<RefCell<TreeNode>>>,\n// }\n// \n// impl TreeNode {\n//   #[inline]\n//   pub fn new(val: i32) -> Self {\n//     TreeNode {\n//       val,\n//       left: None,\n//       right: None\n//     }\n//   }\n// }\nuse std::rc::Rc;\nuse std::cell::RefCell;\nimpl Solution {\n    pub fn vertical_traversal(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<Vec<i32>> {\n        \n    }\n}",
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