{"id":"579c7625-f436-4e53-8e0c-1b8aad7d8964","name":"Validate Bst","description":"1. You are given a partially written function to solve.\r\n2. Determine if it is a valid binary search tree (BST).\r\n\r\n3. A valid BST is defined as follows:\r\n 3.1 The left subtree of a node contains only nodes with keys less than the node's key.\r\n 3.2 The right subtree of a node contains only nodes with keys greater than the node's key.\r\n 3.3 Both the left and right subtrees must also be binary search trees.\r\n","inputFormat":"Input is managed for you.\r\n","outputFormat":"output is managed for you.\r\n","constraints":"0 &lt;= Number of Nodes &lt;= 10^9\r\n-10^9 &lt;= value of Node data &lt;= 10^9\r\n","sampleCode":{"cpp":{"code":"#include <iostream>\r\n#include <vector>\r\nusing namespace std;\r\n\r\nclass TreeNode\r\n{\r\npublic:\r\n int val = 0;\r\n TreeNode *left = nullptr;\r\n TreeNode *right = nullptr;\r\n\r\n TreeNode(int val)\r\n {\r\n this->val = val;\r\n }\r\n};\r\n\r\nbool isValidBST(TreeNode *root)\r\n{\r\n return true;\r\n}\r\n\r\nTreeNode *createTree(vector<int> &arr, vector<int> &IDX)\r\n{\r\n if (IDX[0] > arr.size() || arr[IDX[0]] == -1){\r\n IDX[0]++;\r\n return nullptr;\r\n }\r\n\r\n TreeNode *node = new TreeNode(arr[IDX[0]++]);\r\n node->left = createTree(arr, IDX);\r\n node->right = createTree(arr, IDX);\r\n\r\n return node;\r\n}\r\n\r\nvoid solve()\r\n{\r\n int n;\r\n cin >> n;\r\n vector<int> arr(n, 0);\r\n for (int i = 0; i < n; i++)\r\n {\r\n cin >> arr[i];\r\n }\r\n\r\n vector<int> IDX(1, 0);\r\n TreeNode *root = createTree(arr, IDX);\r\n cout << (boolalpha) << isValidBST(root);\r\n}\r\n\r\nint main()\r\n{\r\n solve();\r\n return 0;\r\n}"},"java":{"code":"import java.util.Scanner;\r\n\r\npublic class Main {\r\n public static Scanner scn = new Scanner(System.in);\r\n\r\n public static class TreeNode {\r\n int val = 0;\r\n TreeNode left = null;\r\n TreeNode right = null;\r\n\r\n TreeNode(int val) {\r\n this.val = val;\r\n }\r\n }\r\n\r\n public static boolean isValidBST(TreeNode root) {\r\n return true;\r\n }\r\n\r\n public static TreeNode createTree(int[] arr, int[] IDX) {\r\n if (IDX[0] > arr.length || arr[IDX[0]] == -1){\r\n IDX[0]++;\r\n return null;\r\n }\r\n\r\n TreeNode node = new TreeNode(arr[IDX[0]++]);\r\n node.left = createTree(arr, IDX);\r\n node.right = createTree(arr, IDX);\r\n\r\n return node;\r\n }\r\n\r\n public static void solve() {\r\n int n = scn.nextInt();\r\n int[] arr = new int[n];\r\n for (int i = 0; i < n; i++)\r\n arr[i] = scn.nextInt();\r\n\r\n int[] IDX = new int[1];\r\n TreeNode root = createTree(arr, IDX);\r\n System.out.println(isValidBST(root));\r\n }\r\n\r\n public static void main(String[] args) {\r\n solve();\r\n }\r\n}"},"python":{"code":""}},"points":10,"difficulty":"easy","sampleInput":"13\r\n7\r\n3\r\n2\r\n-1\r\n-1\r\n5\r\n-1\r\n-1\r\n10\r\n-1\r\n12\r\n-1\r\n-1","sampleOutput":"true\r\n","questionVideo":"https://www.youtube.com/embed/klXjnz8zn2E","hints":[],"associated":[],"solutionSeen":false,"tags":[],"meta":{"path":[{"id":0,"name":"home"},{"id":"0c54b191-7b99-4f2c-acb3-e7f2ec748b2a","name":"Data Structures and Algorithms","slug":"data-structures-and-algorithms","type":0},{"id":"2e9df04c-be14-441c-9a18-8b5a8ca6596d","name":"Trees For Intermediate","slug":"trees-for-intermediate-9994","type":0},{"id":"2a7df609-1502-4138-8b84-315374311ca8","name":"Validate Bst","slug":"validate-bst","type":1}],"next":{"id":"deecfc87-e938-4baf-8b50-6c86f0a2ca72","name":"Validate BST","type":3,"slug":"validate-bst"},"prev":{"id":"852aca4b-685b-45ab-b1e3-2e627af12846","name":"House Robber In Binary Tree","type":0,"slug":"house-robber-in-binary-tree"}}}

Validate Bst

1. You are given a partially written function to solve. 2. Determine if it is a valid binary search tree (BST). 3. A valid BST is defined as follows: 3.1 The left subtree of a node contains only nodes with keys less than the node's key. 3.2 The right subtree of a node contains only nodes with keys greater than the node's key. 3.3 Both the left and right subtrees must also be binary search trees.

{"id":"579c7625-f436-4e53-8e0c-1b8aad7d8964","name":"Validate Bst","description":"1. You are given a partially written function to solve.\r\n2. Determine if it is a valid binary search tree (BST).\r\n\r\n3. A valid BST is defined as follows:\r\n 3.1 The left subtree of a node contains only nodes with keys less than the node's key.\r\n 3.2 The right subtree of a node contains only nodes with keys greater than the node's key.\r\n 3.3 Both the left and right subtrees must also be binary search trees.\r\n","inputFormat":"Input is managed for you.\r\n","outputFormat":"output is managed for you.\r\n","constraints":"0 &lt;= Number of Nodes &lt;= 10^9\r\n-10^9 &lt;= value of Node data &lt;= 10^9\r\n","sampleCode":{"cpp":{"code":"#include <iostream>\r\n#include <vector>\r\nusing namespace std;\r\n\r\nclass TreeNode\r\n{\r\npublic:\r\n int val = 0;\r\n TreeNode *left = nullptr;\r\n TreeNode *right = nullptr;\r\n\r\n TreeNode(int val)\r\n {\r\n this->val = val;\r\n }\r\n};\r\n\r\nbool isValidBST(TreeNode *root)\r\n{\r\n return true;\r\n}\r\n\r\nTreeNode *createTree(vector<int> &arr, vector<int> &IDX)\r\n{\r\n if (IDX[0] > arr.size() || arr[IDX[0]] == -1){\r\n IDX[0]++;\r\n return nullptr;\r\n }\r\n\r\n TreeNode *node = new TreeNode(arr[IDX[0]++]);\r\n node->left = createTree(arr, IDX);\r\n node->right = createTree(arr, IDX);\r\n\r\n return node;\r\n}\r\n\r\nvoid solve()\r\n{\r\n int n;\r\n cin >> n;\r\n vector<int> arr(n, 0);\r\n for (int i = 0; i < n; i++)\r\n {\r\n cin >> arr[i];\r\n }\r\n\r\n vector<int> IDX(1, 0);\r\n TreeNode *root = createTree(arr, IDX);\r\n cout << (boolalpha) << isValidBST(root);\r\n}\r\n\r\nint main()\r\n{\r\n solve();\r\n return 0;\r\n}"},"java":{"code":"import java.util.Scanner;\r\n\r\npublic class Main {\r\n public static Scanner scn = new Scanner(System.in);\r\n\r\n public static class TreeNode {\r\n int val = 0;\r\n TreeNode left = null;\r\n TreeNode right = null;\r\n\r\n TreeNode(int val) {\r\n this.val = val;\r\n }\r\n }\r\n\r\n public static boolean isValidBST(TreeNode root) {\r\n return true;\r\n }\r\n\r\n public static TreeNode createTree(int[] arr, int[] IDX) {\r\n if (IDX[0] > arr.length || arr[IDX[0]] == -1){\r\n IDX[0]++;\r\n return null;\r\n }\r\n\r\n TreeNode node = new TreeNode(arr[IDX[0]++]);\r\n node.left = createTree(arr, IDX);\r\n node.right = createTree(arr, IDX);\r\n\r\n return node;\r\n }\r\n\r\n public static void solve() {\r\n int n = scn.nextInt();\r\n int[] arr = new int[n];\r\n for (int i = 0; i < n; i++)\r\n arr[i] = scn.nextInt();\r\n\r\n int[] IDX = new int[1];\r\n TreeNode root = createTree(arr, IDX);\r\n System.out.println(isValidBST(root));\r\n }\r\n\r\n public static void main(String[] args) {\r\n solve();\r\n }\r\n}"},"python":{"code":""}},"points":10,"difficulty":"easy","sampleInput":"13\r\n7\r\n3\r\n2\r\n-1\r\n-1\r\n5\r\n-1\r\n-1\r\n10\r\n-1\r\n12\r\n-1\r\n-1","sampleOutput":"true\r\n","questionVideo":"https://www.youtube.com/embed/klXjnz8zn2E","hints":[],"associated":[],"solutionSeen":false,"tags":[],"meta":{"path":[{"id":0,"name":"home"},{"id":"0c54b191-7b99-4f2c-acb3-e7f2ec748b2a","name":"Data Structures and Algorithms","slug":"data-structures-and-algorithms","type":0},{"id":"2e9df04c-be14-441c-9a18-8b5a8ca6596d","name":"Trees For Intermediate","slug":"trees-for-intermediate-9994","type":0},{"id":"2a7df609-1502-4138-8b84-315374311ca8","name":"Validate Bst","slug":"validate-bst","type":1}],"next":{"id":"deecfc87-e938-4baf-8b50-6c86f0a2ca72","name":"Validate BST","type":3,"slug":"validate-bst"},"prev":{"id":"852aca4b-685b-45ab-b1e3-2e627af12846","name":"House Robber In Binary Tree","type":0,"slug":"house-robber-in-binary-tree"}}}
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Validate Bst

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1. You are given a partially written function to solve. 2. Determine if it is a valid binary search tree (BST). 3. A valid BST is defined as follows: 3.1 The left subtree of a node contains only nodes with keys less than the node's key. 3.2 The right subtree of a node contains only nodes with keys greater than the node's key. 3.3 Both the left and right subtrees must also be binary search trees.

Constraints

0 <= Number of Nodes <= 10^9 -10^9 <= value of Node data <= 10^9

Format

Input

Input is managed for you.

Output

output is managed for you.

Example

Sample Input

13 7 3 2 -1 -1 5 -1 -1 10 -1 12 -1 -1

Sample Output

true

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