{"id":"ad84cb58-7583-42ee-9c51-015851bd5d5b","name":"Diagonal Order (anti-clock Wise) Of A Binarytree","description":"1. Given a Binary Tree, print Diagonal Order of it anti-clock wise. \r\n2. For more Information watch given video link below.\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;= 10^5\r\n-1000 &lt;= value of Node data &lt;= 1000\r\n","sampleCode":{"cpp":{"code":"#include <iostream>\r\n#include <vector>\r\n#include <queue>\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\nvector<vector<int>> diagonalOrder(TreeNode *root)\r\n{\r\n queue<TreeNode *> que;\r\n que.push(root);\r\n vector<vector<int>> ans;\r\n while (que.size() != 0)\r\n {\r\n int size = que.size();\r\n TreeNode *node = que.front();\r\n que.pop();\r\n\r\n vector<int> smallAns;\r\n\r\n while (node != nullptr)\r\n {\r\n smallAns.push_back(node->val);\r\n if (node->right != nullptr)\r\n { // add in que for next diagonal process.\r\n que.push(node->right);\r\n }\r\n node = node->left; // move forward in respective diagonal.\r\n }\r\n ans.push_back(smallAns);\r\n }\r\n\r\n return ans;\r\n}\r\n\r\n// input_section=================================================\r\n\r\nTreeNode *createTree(vector<int> &arr, vector<int> &IDX)\r\n{\r\n\r\n if (IDX[0] > arr.size() || arr[IDX[0]] == -1)\r\n {\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\r\n vector<vector<int>> ans = diagonalOrder(root);\r\n int idx = 0;\r\n for (vector<int> &i : ans)\r\n {\r\n cout << idx++ << \" -> \";\r\n for (int j : i)\r\n {\r\n cout << j << \" \";\r\n }\r\n cout << endl;\r\n }\r\n}\r\n\r\nint main()\r\n{\r\n solve();\r\n return 0;\r\n}"},"java":{"code":"import java.util.*;\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 ArrayList<ArrayList<Integer>> diagonalOrder(TreeNode root) {\r\n return ans;\r\n }\r\n\r\n // input_section=================================================\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 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\r\n ArrayList<ArrayList<Integer>> ans = diagonalOrder(root);\r\n int idx = 0;\r\n for (ArrayList<Integer> i : ans) {\r\n System.out.print(idx++ + \" -> \");\r\n for (Integer j : i)\r\n System.out.print(j + \" \");\r\n System.out.println();\r\n }\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":"15\r\n1\r\n1\r\n-1\r\n1\r\n1\r\n-1\r\n1\r\n-1\r\n-1\r\n1\r\n-1\r\n-1\r\n1\r\n-1\r\n-1","sampleOutput":"0 -> 1 1 \r\n1 -> 1 1 1 \r\n2 -> 1 1 \r\n","questionVideo":"https://www.youtube.com/embed/BZ49ZTWUbmg","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":"2b6b68c4-4ec5-4666-ba85-524940d38e0e","name":"Diagonal Order (anti-clock Wise) Of A Binarytree","slug":"diagonal-order-anti-clock-wise-of-a-binarytree","type":1}],"next":{"id":"fb4d4c9c-f0d5-40af-b416-c25e74206825","name":"Diagonal Order (anti-clock Wise) Of A Binarytree","type":0,"slug":"diagonal-order-anti-clock-wise-of-a-binarytree"},"prev":{"id":"36216277-9643-4f04-9fd3-65c5dfce20f9","name":"Diagonal order of a binary tree","type":3,"slug":"diagonal-order-of-a-binary-tree"}}}

Diagonal Order (anti-clock Wise) Of A Binarytree

1. Given a Binary Tree, print Diagonal Order of it anti-clock wise. 2. For more Information watch given video link below.

{"id":"ad84cb58-7583-42ee-9c51-015851bd5d5b","name":"Diagonal Order (anti-clock Wise) Of A Binarytree","description":"1. Given a Binary Tree, print Diagonal Order of it anti-clock wise. \r\n2. For more Information watch given video link below.\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;= 10^5\r\n-1000 &lt;= value of Node data &lt;= 1000\r\n","sampleCode":{"cpp":{"code":"#include <iostream>\r\n#include <vector>\r\n#include <queue>\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\nvector<vector<int>> diagonalOrder(TreeNode *root)\r\n{\r\n queue<TreeNode *> que;\r\n que.push(root);\r\n vector<vector<int>> ans;\r\n while (que.size() != 0)\r\n {\r\n int size = que.size();\r\n TreeNode *node = que.front();\r\n que.pop();\r\n\r\n vector<int> smallAns;\r\n\r\n while (node != nullptr)\r\n {\r\n smallAns.push_back(node->val);\r\n if (node->right != nullptr)\r\n { // add in que for next diagonal process.\r\n que.push(node->right);\r\n }\r\n node = node->left; // move forward in respective diagonal.\r\n }\r\n ans.push_back(smallAns);\r\n }\r\n\r\n return ans;\r\n}\r\n\r\n// input_section=================================================\r\n\r\nTreeNode *createTree(vector<int> &arr, vector<int> &IDX)\r\n{\r\n\r\n if (IDX[0] > arr.size() || arr[IDX[0]] == -1)\r\n {\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\r\n vector<vector<int>> ans = diagonalOrder(root);\r\n int idx = 0;\r\n for (vector<int> &i : ans)\r\n {\r\n cout << idx++ << \" -> \";\r\n for (int j : i)\r\n {\r\n cout << j << \" \";\r\n }\r\n cout << endl;\r\n }\r\n}\r\n\r\nint main()\r\n{\r\n solve();\r\n return 0;\r\n}"},"java":{"code":"import java.util.*;\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 ArrayList<ArrayList<Integer>> diagonalOrder(TreeNode root) {\r\n return ans;\r\n }\r\n\r\n // input_section=================================================\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 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\r\n ArrayList<ArrayList<Integer>> ans = diagonalOrder(root);\r\n int idx = 0;\r\n for (ArrayList<Integer> i : ans) {\r\n System.out.print(idx++ + \" -> \");\r\n for (Integer j : i)\r\n System.out.print(j + \" \");\r\n System.out.println();\r\n }\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":"15\r\n1\r\n1\r\n-1\r\n1\r\n1\r\n-1\r\n1\r\n-1\r\n-1\r\n1\r\n-1\r\n-1\r\n1\r\n-1\r\n-1","sampleOutput":"0 -> 1 1 \r\n1 -> 1 1 1 \r\n2 -> 1 1 \r\n","questionVideo":"https://www.youtube.com/embed/BZ49ZTWUbmg","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":"2b6b68c4-4ec5-4666-ba85-524940d38e0e","name":"Diagonal Order (anti-clock Wise) Of A Binarytree","slug":"diagonal-order-anti-clock-wise-of-a-binarytree","type":1}],"next":{"id":"fb4d4c9c-f0d5-40af-b416-c25e74206825","name":"Diagonal Order (anti-clock Wise) Of A Binarytree","type":0,"slug":"diagonal-order-anti-clock-wise-of-a-binarytree"},"prev":{"id":"36216277-9643-4f04-9fd3-65c5dfce20f9","name":"Diagonal order of a binary tree","type":3,"slug":"diagonal-order-of-a-binary-tree"}}}
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Diagonal Order (anti-clock Wise) Of A Binarytree

easy

1. Given a Binary Tree, print Diagonal Order of it anti-clock wise. 2. For more Information watch given video link below.

Constraints

0 <= Number of Nodes <= 10^5 -1000 <= value of Node data <= 1000

Format

Input

Input is managed for you.

Output

Output is managed for you.

Example

Sample Input

15 1 1 -1 1 1 -1 1 -1 -1 1 -1 -1 1 -1 -1

Sample Output

0 -> 1 1 1 -> 1 1 1 2 -> 1 1

Question Video

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