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{
"data": {
"question": {
"questionId": "107",
"questionFrontendId": "107",
"categoryTitle": "Algorithms",
"boundTopicId": 1314,
"title": "Binary Tree Level Order Traversal II",
"titleSlug": "binary-tree-level-order-traversal-ii",
"content": "<p>Given the <code>root</code> of a binary tree, return <em>the bottom-up level order traversal of its nodes&#39; values</em>. (i.e., from left to right, level by level from leaf to root).</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/02/19/tree1.jpg\" style=\"width: 277px; height: 302px;\" />\n<pre>\n<strong>Input:</strong> root = [3,9,20,null,null,15,7]\n<strong>Output:</strong> [[15,7],[9,20],[3]]\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> root = [1]\n<strong>Output:</strong> [[1]]\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> root = []\n<strong>Output:</strong> []\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>[0, 2000]</code>.</li>\n\t<li><code>-1000 &lt;= Node.val &lt;= 1000</code></li>\n</ul>\n",
"translatedTitle": "二叉树的层序遍历 II",
"translatedContent": "<p>给你二叉树的根节点 <code>root</code> ,返回其节点值 <strong>自底向上的层序遍历</strong> 。 (即按从叶子节点所在层到根节点所在的层,逐层从左向右遍历)</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/02/19/tree1.jpg\" style=\"width: 277px; height: 302px;\" />\n<pre>\n<strong>输入:</strong>root = [3,9,20,null,null,15,7]\n<strong>输出:</strong>[[15,7],[9,20],[3]]\n</pre>\n\n<p><strong>示例 2</strong></p>\n\n<pre>\n<strong>输入:</strong>root = [1]\n<strong>输出:</strong>[[1]]\n</pre>\n\n<p><strong>示例 3</strong></p>\n\n<pre>\n<strong>输入:</strong>root = []\n<strong>输出:</strong>[]\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li>树中节点数目在范围 <code>[0, 2000]</code> 内</li>\n\t<li><code>-1000 &lt;= Node.val &lt;= 1000</code></li>\n</ul>\n",
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"name": "Tree",
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"name": "Breadth-First Search",
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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>> levelOrderBottom(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>> levelOrderBottom(TreeNode root) {\n\n }\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 * 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** levelOrderBottom(struct TreeNode* root, int* returnSize, int** returnColumnSizes) {\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 levelOrderBottom = function(root) {\n\n};",
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"code": "/**\n * Definition for a binary tree node.\n * class TreeNode {\n * val: number\n * left: TreeNode | null\n * right: TreeNode | null\n * constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {\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\nfunction levelOrderBottom(root: TreeNode | null): number[][] {\n \n};",
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"lang": "Swift",
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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 levelOrderBottom(_ root: TreeNode?) -> [[Int]] {\n\n }\n}",
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"lang": "Kotlin",
"langSlug": "kotlin",
"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 levelOrderBottom(root: TreeNode?): List<List<Int>> {\n\n }\n}",
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"code": "/**\n * Definition for a binary tree node.\n * class TreeNode {\n * int val;\n * TreeNode? left;\n * TreeNode? right;\n * TreeNode([this.val = 0, this.left, this.right]);\n * }\n */\nclass Solution {\n List<List<int>> levelOrderBottom(TreeNode? root) {\n \n }\n}",
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"lang": "Go",
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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 levelOrderBottom(root *TreeNode) [][]int {\n\n}",
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"lang": "Ruby",
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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 level_order_bottom(root)\n\nend",
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"code": "/**\n * Definition for a binary tree node.\n * class TreeNode(_value: Int = 0, _left: TreeNode = null, _right: TreeNode = null) {\n * var value: Int = _value\n * var left: TreeNode = _left\n * var right: TreeNode = _right\n * }\n */\nobject Solution {\n def levelOrderBottom(root: TreeNode): List[List[Int]] = {\n\n }\n}",
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"lang": "Racket",
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"code": "; Definition for a binary tree node.\n#|\n\n; val : integer?\n; left : (or/c tree-node? #f)\n; right : (or/c tree-node? #f)\n(struct tree-node\n (val left right) #:mutable #:transparent)\n\n; constructor\n(define (make-tree-node [val 0])\n (tree-node val #f #f))\n\n|#\n\n(define/contract (level-order-bottom root)\n (-> (or/c tree-node? #f) (listof (listof exact-integer?)))\n )",
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"code": "%% Definition for a binary tree node.\n%%\n%% -record(tree_node, {val = 0 :: integer(),\n%% left = null :: 'null' | #tree_node{},\n%% right = null :: 'null' | #tree_node{}}).\n\n-spec level_order_bottom(Root :: #tree_node{} | null) -> [[integer()]].\nlevel_order_bottom(Root) ->\n .",
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"code": "# Definition for a binary tree node.\n#\n# defmodule TreeNode do\n# @type t :: %__MODULE__{\n# val: integer,\n# left: TreeNode.t() | nil,\n# right: TreeNode.t() | nil\n# }\n# defstruct val: 0, left: nil, right: nil\n# end\n\ndefmodule Solution do\n @spec level_order_bottom(root :: TreeNode.t | nil) :: [[integer]]\n def level_order_bottom(root) do\n \n end\nend",
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