{"id":"23306c6d-8637-4a74-b8c3-0422e8653435","name":"Diameter Of A Binary Tree","description":"1. You are given a partially written BinaryTree class.\r\n2. You are required to complete the body of diameter1 function. The function is expected to return the number of edges between two nodes which are farthest from each other in terms of edges.\r\n3. Input and Output is managed for you.","inputFormat":"Input is managed for you.","outputFormat":"Output is managed for you.","constraints":"None","sampleCode":{"cpp":{"code":"#include <iostream>\n#include <vector>\n\nusing namespace std;\n\nclass Node\n{\npublic:\n int data=0;\n Node *left = nullptr;\n Node *right = nullptr;\n Node(int data)\n {\n this->data = data;\n }\n};\n\n class Pair {\n public:\n Node *node=nullptr;\n int state=0;\n\n Pair(Node *node, int state) {\n this->node = node;\n this->state = state;\n }\n };\n\nint idx = 0;\nNode *constructTree(vector<int> &arr)\n{\n\n if (idx == arr.size() || arr[idx] == -1)\n {\n idx++;\n return nullptr;\n }\n Node *node = new Node(arr[idx++]);\n node->left = constructTree(arr);\n node->right = constructTree(arr);\n return node;\n}\n\n//Display function\nvoid display(Node *node)\n{\n if (node == nullptr)\n return;\n string str = \"\";\n str += node->left != nullptr ? to_string(node->left->data) : \".\";\n str += \" <- \" + to_string(node->data) + \" -> \";\n str += node->right != nullptr ? to_string(node->right->data) : \".\";\n cout << str << endl;\n\n display(node->left);\n display(node->right);\n}\n\n\n//Height function\nint height(Node *node)\n{\n return node == nullptr ? -1 : max(height(node->left), height(node->right)) + 1; // for no of edges: -1, and in terms of no of nodes return 0;\n}\n\n\nint diameter(Node *root)\n{\n // write your code here \n}\n\nint main(){\n int n;\n cin>>n;\n \n vector<int> arr(n,0);\n for(int i = 0; i < n; i++) {\n string tmp;\n cin>>tmp;\n if (tmp==\"n\") {\n arr[i] = -1;\n } else {\n arr[i] = stoi(tmp);\n }\n }\n \n \n Node * root = constructTree(arr);\n\n int dia = 0;\n dia = diameter(root);\n cout<<dia;\n}"},"java":{"code":"import java.io.*;\r\nimport java.util.*;\r\n\r\npublic class Main {\r\n public static class Node {\r\n int data;\r\n Node left;\r\n Node right;\r\n\r\n Node(int data, Node left, Node right) {\r\n this.data = data;\r\n this.left = left;\r\n this.right = right;\r\n }\r\n }\r\n\r\n public static class Pair {\r\n Node node;\r\n int state;\r\n\r\n Pair(Node node, int state) {\r\n this.node = node;\r\n this.state = state;\r\n }\r\n }\r\n\r\n public static Node construct(Integer[] arr) {\r\n Node root = new Node(arr[0], null, null);\r\n Pair rtp = new Pair(root, 1);\r\n\r\n Stack<Pair> st = new Stack<>();\r\n st.push(rtp);\r\n\r\n int idx = 0;\r\n while (st.size() > 0) {\r\n Pair top = st.peek();\r\n if (top.state == 1) {\r\n idx++;\r\n if (arr[idx] != null) {\r\n top.node.left = new Node(arr[idx], null, null);\r\n Pair lp = new Pair(top.node.left, 1);\r\n st.push(lp);\r\n } else {\r\n top.node.left = null;\r\n }\r\n\r\n top.state++;\r\n } else if (top.state == 2) {\r\n idx++;\r\n if (arr[idx] != null) {\r\n top.node.right = new Node(arr[idx], null, null);\r\n Pair rp = new Pair(top.node.right, 1);\r\n st.push(rp);\r\n } else {\r\n top.node.right = null;\r\n }\r\n\r\n top.state++;\r\n } else {\r\n st.pop();\r\n }\r\n }\r\n\r\n return root;\r\n }\r\n\r\n public static void display(Node node) {\r\n if (node == null) {\r\n return;\r\n }\r\n\r\n String str = \"\";\r\n str += node.left == null ? \".\" : node.left.data + \"\";\r\n str += \" <- \" + node.data + \" -> \";\r\n str += node.right == null ? \".\" : node.right.data + \"\";\r\n System.out.println(str);\r\n\r\n display(node.left);\r\n display(node.right);\r\n }\r\n\r\n public static int height(Node node) {\r\n if (node == null) {\r\n return -1;\r\n }\r\n\r\n int lh = height(node.left);\r\n int rh = height(node.right);\r\n\r\n int th = Math.max(lh, rh) + 1;\r\n return th;\r\n }\r\n\r\n public static int diameter1(Node node) {\r\n // write your code here\r\n }\r\n\r\n public static void main(String[] args) throws Exception {\r\n BufferedReader br = new BufferedReader(new InputStreamReader(System.in));\r\n int n = Integer.parseInt(br.readLine());\r\n Integer[] arr = new Integer[n];\r\n String[] values = br.readLine().split(\" \");\r\n for (int i = 0; i < n; i++) {\r\n if (values[i].equals(\"n\") == false) {\r\n arr[i] = Integer.parseInt(values[i]);\r\n } else {\r\n arr[i] = null;\r\n }\r\n }\r\n\r\n Node root = construct(arr);\r\n\r\n int diameter = 0;\r\n diameter = diameter1(root);\r\n System.out.println(diameter);\r\n }\r\n\r\n}"},"python":{"code":"class Node:\n \n def __init__(self,data,left,right):\n self.data = data\n self.left = None\n self.right = None\nclass Pair:\n def __init__(self,node,state):\n self.node = node\n self.state = state\n\ndef construct(arr):\n root=Node(arr[0],None,None)\n rtp=Pair(root,1);\n \n st=[]\n st.append(rtp);\n \n idx=0;\n n = len(arr)\n while(len(st)>0):\n top=st[-1];\n if top.state==1:\n idx+=1\n st[-1].state+=1\n if(arr[idx]!=-1):\n top.node.left = Node(arr[idx], None, None);\n lp = Pair(top.node.left, 1);\n st.append(lp);\n else:\n top.node.left=None;\n elif top.state==2:\n idx+=1\n st[-1].state+=1\n if(arr[idx]!=-1):\n top.node.right = Node(arr[idx], None, None);\n rp = Pair(top.node.right, 1);\n st.append(rp);\n else:\n top.node.right=None;\n else:\n st.pop()\n return root;\n\ndef height(node):\n if(node == None):\n return -1\n else:\n ans = max(height(node.left),height(node.right))+1\n return ans\n\n \ndef diameter(node):\n # write your code here\n \n\nn = int(input())\nst = input()\narr = [0]*n\narr = list(map(int,st.replace(\"n\",\"-1\").split(\" \")));\n\n\nroot = construct(arr)\nr = diameter(root)\nprint(r)"}},"points":10,"difficulty":"easy","sampleInput":"19\r\n50 25 12 n n 37 30 n n n 75 62 n 70 n n 87 n n","sampleOutput":"6","questionVideo":"https://www.youtube.com/embed/aULnlLj1LmU","hints":[],"associated":[{"id":"6be780c8-d7bb-43af-b4bf-16ca5961461b","name":"MCQ4 :","slug":"mcq4-kb7rm","type":4},{"id":"9ad89c01-7b3e-4ed1-aeed-49b3668a5c7e","name":"what is the time complexity of the optimized approach of the solution?","slug":"what-is-the-time-complexity-of-the-optimized-approach-of-the-solution","type":4},{"id":"f1384e32-75ee-46e6-9ed1-c052f5fc3a33","name":"is it mandatory the diameter of a tree must passed fron the root node.","slug":"is-it-mandatory-the-diameter-of-a-tree-must-passed-fron-the-root-node","type":4},{"id":"fabc754d-309c-4cf0-a26f-bb8b14a47f76","name":"what is the formula of diameter of a binary tree","slug":"what-is-the-formula-of-diameter-of-a-binary-tree","type":4}],"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":"b042fe97-356f-41ed-80fd-abacd61fc234","name":"Binary Tree For Beginners","slug":"binary-tree-for-beginners","type":0},{"id":"74253ae7-e696-4a4b-b6ad-b836b05186b6","name":"Diameter Of A Binary Tree","slug":"diameter-of-a-binary-tree","type":1}],"next":{"id":"60672207-6bbd-4fcf-bf2e-05d4dd071f63","name":"Diameter of a Binary Tree","type":3,"slug":"diameter-of-a-binary-tree"},"prev":{"id":"17a1ed44-1063-44ba-9867-0f6cd63a82a6","name":"Remove Leaves In Binary Tree","type":3,"slug":"remove-leaves-in-binary-tree"}}}

Diameter Of A Binary Tree

1. You are given a partially written BinaryTree class. 2. You are required to complete the body of diameter1 function. The function is expected to return the number of edges between two nodes which are farthest from each other in terms of edges. 3. Input and Output is managed for you.

{"id":"23306c6d-8637-4a74-b8c3-0422e8653435","name":"Diameter Of A Binary Tree","description":"1. You are given a partially written BinaryTree class.\r\n2. You are required to complete the body of diameter1 function. The function is expected to return the number of edges between two nodes which are farthest from each other in terms of edges.\r\n3. Input and Output is managed for you.","inputFormat":"Input is managed for you.","outputFormat":"Output is managed for you.","constraints":"None","sampleCode":{"cpp":{"code":"#include <iostream>\n#include <vector>\n\nusing namespace std;\n\nclass Node\n{\npublic:\n int data=0;\n Node *left = nullptr;\n Node *right = nullptr;\n Node(int data)\n {\n this->data = data;\n }\n};\n\n class Pair {\n public:\n Node *node=nullptr;\n int state=0;\n\n Pair(Node *node, int state) {\n this->node = node;\n this->state = state;\n }\n };\n\nint idx = 0;\nNode *constructTree(vector<int> &arr)\n{\n\n if (idx == arr.size() || arr[idx] == -1)\n {\n idx++;\n return nullptr;\n }\n Node *node = new Node(arr[idx++]);\n node->left = constructTree(arr);\n node->right = constructTree(arr);\n return node;\n}\n\n//Display function\nvoid display(Node *node)\n{\n if (node == nullptr)\n return;\n string str = \"\";\n str += node->left != nullptr ? to_string(node->left->data) : \".\";\n str += \" <- \" + to_string(node->data) + \" -> \";\n str += node->right != nullptr ? to_string(node->right->data) : \".\";\n cout << str << endl;\n\n display(node->left);\n display(node->right);\n}\n\n\n//Height function\nint height(Node *node)\n{\n return node == nullptr ? -1 : max(height(node->left), height(node->right)) + 1; // for no of edges: -1, and in terms of no of nodes return 0;\n}\n\n\nint diameter(Node *root)\n{\n // write your code here \n}\n\nint main(){\n int n;\n cin>>n;\n \n vector<int> arr(n,0);\n for(int i = 0; i < n; i++) {\n string tmp;\n cin>>tmp;\n if (tmp==\"n\") {\n arr[i] = -1;\n } else {\n arr[i] = stoi(tmp);\n }\n }\n \n \n Node * root = constructTree(arr);\n\n int dia = 0;\n dia = diameter(root);\n cout<<dia;\n}"},"java":{"code":"import java.io.*;\r\nimport java.util.*;\r\n\r\npublic class Main {\r\n public static class Node {\r\n int data;\r\n Node left;\r\n Node right;\r\n\r\n Node(int data, Node left, Node right) {\r\n this.data = data;\r\n this.left = left;\r\n this.right = right;\r\n }\r\n }\r\n\r\n public static class Pair {\r\n Node node;\r\n int state;\r\n\r\n Pair(Node node, int state) {\r\n this.node = node;\r\n this.state = state;\r\n }\r\n }\r\n\r\n public static Node construct(Integer[] arr) {\r\n Node root = new Node(arr[0], null, null);\r\n Pair rtp = new Pair(root, 1);\r\n\r\n Stack<Pair> st = new Stack<>();\r\n st.push(rtp);\r\n\r\n int idx = 0;\r\n while (st.size() > 0) {\r\n Pair top = st.peek();\r\n if (top.state == 1) {\r\n idx++;\r\n if (arr[idx] != null) {\r\n top.node.left = new Node(arr[idx], null, null);\r\n Pair lp = new Pair(top.node.left, 1);\r\n st.push(lp);\r\n } else {\r\n top.node.left = null;\r\n }\r\n\r\n top.state++;\r\n } else if (top.state == 2) {\r\n idx++;\r\n if (arr[idx] != null) {\r\n top.node.right = new Node(arr[idx], null, null);\r\n Pair rp = new Pair(top.node.right, 1);\r\n st.push(rp);\r\n } else {\r\n top.node.right = null;\r\n }\r\n\r\n top.state++;\r\n } else {\r\n st.pop();\r\n }\r\n }\r\n\r\n return root;\r\n }\r\n\r\n public static void display(Node node) {\r\n if (node == null) {\r\n return;\r\n }\r\n\r\n String str = \"\";\r\n str += node.left == null ? \".\" : node.left.data + \"\";\r\n str += \" <- \" + node.data + \" -> \";\r\n str += node.right == null ? \".\" : node.right.data + \"\";\r\n System.out.println(str);\r\n\r\n display(node.left);\r\n display(node.right);\r\n }\r\n\r\n public static int height(Node node) {\r\n if (node == null) {\r\n return -1;\r\n }\r\n\r\n int lh = height(node.left);\r\n int rh = height(node.right);\r\n\r\n int th = Math.max(lh, rh) + 1;\r\n return th;\r\n }\r\n\r\n public static int diameter1(Node node) {\r\n // write your code here\r\n }\r\n\r\n public static void main(String[] args) throws Exception {\r\n BufferedReader br = new BufferedReader(new InputStreamReader(System.in));\r\n int n = Integer.parseInt(br.readLine());\r\n Integer[] arr = new Integer[n];\r\n String[] values = br.readLine().split(\" \");\r\n for (int i = 0; i < n; i++) {\r\n if (values[i].equals(\"n\") == false) {\r\n arr[i] = Integer.parseInt(values[i]);\r\n } else {\r\n arr[i] = null;\r\n }\r\n }\r\n\r\n Node root = construct(arr);\r\n\r\n int diameter = 0;\r\n diameter = diameter1(root);\r\n System.out.println(diameter);\r\n }\r\n\r\n}"},"python":{"code":"class Node:\n \n def __init__(self,data,left,right):\n self.data = data\n self.left = None\n self.right = None\nclass Pair:\n def __init__(self,node,state):\n self.node = node\n self.state = state\n\ndef construct(arr):\n root=Node(arr[0],None,None)\n rtp=Pair(root,1);\n \n st=[]\n st.append(rtp);\n \n idx=0;\n n = len(arr)\n while(len(st)>0):\n top=st[-1];\n if top.state==1:\n idx+=1\n st[-1].state+=1\n if(arr[idx]!=-1):\n top.node.left = Node(arr[idx], None, None);\n lp = Pair(top.node.left, 1);\n st.append(lp);\n else:\n top.node.left=None;\n elif top.state==2:\n idx+=1\n st[-1].state+=1\n if(arr[idx]!=-1):\n top.node.right = Node(arr[idx], None, None);\n rp = Pair(top.node.right, 1);\n st.append(rp);\n else:\n top.node.right=None;\n else:\n st.pop()\n return root;\n\ndef height(node):\n if(node == None):\n return -1\n else:\n ans = max(height(node.left),height(node.right))+1\n return ans\n\n \ndef diameter(node):\n # write your code here\n \n\nn = int(input())\nst = input()\narr = [0]*n\narr = list(map(int,st.replace(\"n\",\"-1\").split(\" \")));\n\n\nroot = construct(arr)\nr = diameter(root)\nprint(r)"}},"points":10,"difficulty":"easy","sampleInput":"19\r\n50 25 12 n n 37 30 n n n 75 62 n 70 n n 87 n n","sampleOutput":"6","questionVideo":"https://www.youtube.com/embed/aULnlLj1LmU","hints":[],"associated":[{"id":"6be780c8-d7bb-43af-b4bf-16ca5961461b","name":"MCQ4 :","slug":"mcq4-kb7rm","type":4},{"id":"9ad89c01-7b3e-4ed1-aeed-49b3668a5c7e","name":"what is the time complexity of the optimized approach of the solution?","slug":"what-is-the-time-complexity-of-the-optimized-approach-of-the-solution","type":4},{"id":"f1384e32-75ee-46e6-9ed1-c052f5fc3a33","name":"is it mandatory the diameter of a tree must passed fron the root node.","slug":"is-it-mandatory-the-diameter-of-a-tree-must-passed-fron-the-root-node","type":4},{"id":"fabc754d-309c-4cf0-a26f-bb8b14a47f76","name":"what is the formula of diameter of a binary tree","slug":"what-is-the-formula-of-diameter-of-a-binary-tree","type":4}],"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":"b042fe97-356f-41ed-80fd-abacd61fc234","name":"Binary Tree For Beginners","slug":"binary-tree-for-beginners","type":0},{"id":"74253ae7-e696-4a4b-b6ad-b836b05186b6","name":"Diameter Of A Binary Tree","slug":"diameter-of-a-binary-tree","type":1}],"next":{"id":"60672207-6bbd-4fcf-bf2e-05d4dd071f63","name":"Diameter of a Binary Tree","type":3,"slug":"diameter-of-a-binary-tree"},"prev":{"id":"17a1ed44-1063-44ba-9867-0f6cd63a82a6","name":"Remove Leaves In Binary Tree","type":3,"slug":"remove-leaves-in-binary-tree"}}}
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Diameter Of A Binary Tree

easy

1. You are given a partially written BinaryTree class. 2. You are required to complete the body of diameter1 function. The function is expected to return the number of edges between two nodes which are farthest from each other in terms of edges. 3. Input and Output is managed for you.

Constraints

None

Format

Input

Input is managed for you.

Output

Output is managed for you.

Example

Sample Input

19 50 25 12 n n 37 30 n n n 75 62 n 70 n n 87 n n

Sample Output

6

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