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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,42 @@ | ||
| package com_github_leetcode | ||
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| import scala.collection.mutable | ||
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| class TreeNode(var value: Int, var left: TreeNode = null, var right: TreeNode = null) { | ||
| override def toString: String = { | ||
| if (left == null && right == null) { | ||
| value.toString | ||
| } else { | ||
| val root = value.toString | ||
| val leftValue = Option(left).map(_.toString).getOrElse("null") | ||
| val rightValue = Option(right).map(_.toString).getOrElse("null") | ||
| s"$root,$leftValue,$rightValue" | ||
| } | ||
| } | ||
| } | ||
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| object TreeNode { | ||
| def create(treeValues: List[Option[Int]]): TreeNode = { | ||
| if (treeValues.isEmpty) { | ||
| null | ||
| } else { | ||
| val root = new TreeNode(treeValues.head.getOrElse(0)) | ||
| val queue = mutable.Queue(root) | ||
| var i = 1 | ||
| while (i < treeValues.length) { | ||
| val curr = queue.dequeue() | ||
| if (treeValues(i).isDefined) { | ||
| curr.left = new TreeNode(treeValues(i).get) | ||
| queue.enqueue(curr.left) | ||
| } | ||
| i += 1 | ||
| if (i < treeValues.length && treeValues(i).isDefined) { | ||
| curr.right = new TreeNode(treeValues(i).get) | ||
| queue.enqueue(curr.right) | ||
| } | ||
| i += 1 | ||
| } | ||
| root | ||
| } | ||
| } | ||
| } |
41 changes: 41 additions & 0 deletions
41
src/main/scala/g0001_0100/s0094_binary_tree_inorder_traversal/Solution.scala
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| package g0001_0100.s0094_binary_tree_inorder_traversal | ||
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| // #Easy #Top_100_Liked_Questions #Top_Interview_Questions #Depth_First_Search #Tree #Binary_Tree | ||
| // #Stack #Data_Structure_I_Day_10_Tree #Udemy_Tree_Stack_Queue #Big_O_Time_O(n)_Space_O(n) | ||
| // #2023_11_03_Time_456_ms_(68.42%)_Space_57.3_MB_(7.89%) | ||
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| import com_github_leetcode.TreeNode | ||
| import scala.collection.mutable.ListBuffer | ||
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| /* | ||
| * Definition for a binary tree node. | ||
| * class TreeNode(_value: Int = 0, _left: TreeNode = null, _right: TreeNode = null) { | ||
| * var value: Int = _value | ||
| * var left: TreeNode = _left | ||
| * var right: TreeNode = _right | ||
| * } | ||
| */ | ||
| object Solution { | ||
| def inorderTraversal(root: TreeNode): List[Int] = { | ||
| if (root == null) { | ||
| List.empty | ||
| } else { | ||
| val answer = ListBuffer[Int]() | ||
| inorderTraversal(root, answer) | ||
| answer.toList | ||
| } | ||
| } | ||
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| private def inorderTraversal(root: TreeNode, answer: ListBuffer[Int]): Unit = { | ||
| if (root == null) { | ||
| return | ||
| } | ||
| if (root.left != null) { | ||
| inorderTraversal(root.left, answer) | ||
| } | ||
| answer += root.value | ||
| if (root.right != null) { | ||
| inorderTraversal(root.right, answer) | ||
| } | ||
| } | ||
| } |
48 changes: 48 additions & 0 deletions
48
src/main/scala/g0001_0100/s0094_binary_tree_inorder_traversal/readme.md
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| Original file line number | Diff line number | Diff line change |
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| @@ -0,0 +1,48 @@ | ||
| 94\. Binary Tree Inorder Traversal | ||
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| Easy | ||
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| Given the `root` of a binary tree, return _the inorder traversal of its nodes' values_. | ||
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| **Example 1:** | ||
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|  | ||
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| **Input:** root = [1,null,2,3] | ||
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| **Output:** [1,3,2] | ||
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| **Example 2:** | ||
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| **Input:** root = [] | ||
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| **Output:** [] | ||
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| **Example 3:** | ||
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| **Input:** root = [1] | ||
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| **Output:** [1] | ||
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| **Example 4:** | ||
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|  | ||
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| **Input:** root = [1,2] | ||
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| **Output:** [2,1] | ||
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| **Example 5:** | ||
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|  | ||
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| **Input:** root = [1,null,2] | ||
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| **Output:** [1,2] | ||
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| **Constraints:** | ||
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| * The number of nodes in the tree is in the range `[0, 100]`. | ||
| * `-100 <= Node.val <= 100` | ||
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| **Follow up:** Recursive solution is trivial, could you do it iteratively? |
17 changes: 17 additions & 0 deletions
17
src/main/scala/g0001_0100/s0096_unique_binary_search_trees/Solution.scala
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| package g0001_0100.s0096_unique_binary_search_trees | ||
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| // #Medium #Top_100_Liked_Questions #Dynamic_Programming #Math #Tree #Binary_Tree | ||
| // #Binary_Search_Tree #Dynamic_Programming_I_Day_11 #Big_O_Time_O(n)_Space_O(1) | ||
| // #2023_11_03_Time_403_ms_(66.67%)_Space_51.5_MB_(100.00%) | ||
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| object Solution { | ||
| def numTrees(n: Int): Int = { | ||
| var result: Long = 1L | ||
| for (i <- 0 until n) { | ||
| result *= 2L * n - i | ||
| result /= i + 1 | ||
| } | ||
| result /= (n + 1) | ||
| result.toInt | ||
| } | ||
| } |
23 changes: 23 additions & 0 deletions
23
src/main/scala/g0001_0100/s0096_unique_binary_search_trees/readme.md
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,23 @@ | ||
| 96\. Unique Binary Search Trees | ||
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| Medium | ||
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| Given an integer `n`, return _the number of structurally unique **BST'**s (binary search trees) which has exactly_ `n` _nodes of unique values from_ `1` _to_ `n`. | ||
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| **Example 1:** | ||
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|  | ||
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| **Input:** n = 3 | ||
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| **Output:** 5 | ||
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| **Example 2:** | ||
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| **Input:** n = 1 | ||
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| **Output:** 1 | ||
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| **Constraints:** | ||
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| * `1 <= n <= 19` |
32 changes: 32 additions & 0 deletions
32
src/main/scala/g0001_0100/s0098_validate_binary_search_tree/Solution.scala
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,32 @@ | ||
| package g0001_0100.s0098_validate_binary_search_tree | ||
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| // #Medium #Top_100_Liked_Questions #Top_Interview_Questions #Depth_First_Search #Tree #Binary_Tree | ||
| // #Binary_Search_Tree #Data_Structure_I_Day_14_Tree #Level_1_Day_8_Binary_Search_Tree | ||
| // #Udemy_Tree_Stack_Queue #Big_O_Time_O(N)_Space_O(log(N)) | ||
| // #2023_11_03_Time_507_ms_(70.21%)_Space_57_MB_(93.62%) | ||
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| import com_github_leetcode.TreeNode | ||
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| /* | ||
| * Definition for a binary tree node. | ||
| * class TreeNode(_value: Int = 0, _left: TreeNode = null, _right: TreeNode = null) { | ||
| * var value: Int = _value | ||
| * var left: TreeNode = _left | ||
| * var right: TreeNode = _right | ||
| * } | ||
| */ | ||
| object Solution { | ||
| def isValidBST(root: TreeNode): Boolean = { | ||
| def solve(node: TreeNode, left: Long, right: Long): Boolean = { | ||
| if (node == null) { | ||
| true | ||
| } else if (node.value <= left || node.value >= right) { | ||
| false | ||
| } else { | ||
| solve(node.left, left, node.value) && solve(node.right, node.value, right) | ||
| } | ||
| } | ||
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| solve(root, Long.MinValue, Long.MaxValue) | ||
| } | ||
| } |
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