Verilen bir özel ikili ağaç kimin yaprak düğümleri bir form oluşturmak üzere bağlanırlar dairesel çift bağlantılı liste görev bulmaktır yükseklik ağacın.
Örnekler:
Giriş:
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Çıkış: 2
Açıklama: Yaprak düğümlerini tanıdıktan sonra ikili ağacın yüksekliği 2'dir. Yukarıdaki ikili ağaçta 6, 5 ve 4 yaprak düğümleridir ve dairesel, çift bağlantılı bir liste oluştururlar. Burada yaprak düğümün sol işaretçisi, dairesel çift bağlantılı listenin önceki işaretçisi olarak görev yapacak ve sağ işaretçisi, dairesel çift bağlantılı listenin sonraki işaretçisi olarak görev yapacak.Giriş:
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Çıkış: 1
Açıklama: Yaprak düğümleri tanındıktan sonra ikili ağacın yüksekliği 1'dir. Yukarıdaki ikili ağaçta 2 ve 3 yaprak düğümleridir ve dairesel, çift bağlantılı bir liste oluştururlar.
Yaklaşmak :
C++Fikir takip etmektir benzer yaklaşım bizim için yaptığımız gibi normal bir ikili ağacın yüksekliğini bulma . Biz yinelemeli olarak hesaplamak yükseklik ile ilgili sol ve sağ bir düğümün alt ağaçları ve atama yükseklik olarak düğüme maksimum iki çocuğun boyu artı 1. Ancak bir çocuğun sol ve sağ çocuğu yaprak düğümü normal ikili ağaçlar için boştur. Ancak burada yaprak düğümü dairesel çift bağlantılı bir liste düğümüdür. Yani bir düğümün yaprak düğüm olması için şunu kontrol ederiz: düğümün solu sağı işaret ediyor düğüm ve onun sağ sol aynı zamanda şuna da işaret ediyor düğüm kendisi.
// C++ program to calculate height of a special tree // whose leaf nodes forms a circular doubly linked list #include using namespace std; class Node { public: int data; Node *left *right; Node(int x) { data = x; left = nullptr; right = nullptr; } }; // function to check if given // node is a leaf node or node bool isLeaf(Node* node) { // For a node to be a leaf node it should // satisfy the following two conditions: // 1. Node's left's right pointer should be // current node. // 2. Node's right's left pointer should be // current node. // If one condition is met it is guaranteed // that the other consition is also true. return node->left && node->left->right == node && node->right && node->right->left == node; } // Compute the height of a tree int findTreeHeight(Node* node) { // if node is NULL return -1. if (node == nullptr) return -1; // if node is a leaf node return 0 if (isLeaf(node)) return 0; // compute the depth of each subtree // and take maximum return 1 + max(findTreeHeight(node->left) findTreeHeight(node->right)); } int main() { Node* root = new Node(1); root->left = new Node(2); root->right = new Node(3); root->left->left = new Node(4); root->left->right = new Node(5); root->left->left->left = new Node(6); // Given tree contains 3 leaf nodes Node* l1 = root->left->left->left; Node* l2 = root->left->right; Node* l3 = root->right; // create circular doubly linked list out of // leaf nodes of the tree // set next pointer of linked list l1->right = l2 l2->right = l3 l3->right = l1; // set prev pointer of linked list l3->left = l2 l2->left = l1 l1->left = l3; cout << findTreeHeight(root); return 0; }
C // C program to calculate height of a special tree // whose leaf nodes forms a circular doubly linked list #include #include struct Node { int data; struct Node *left *right; }; // function to check if given // node is a leaf node or node int isLeaf(struct Node* node) { // For a node to be a leaf node it should // satisfy the following two conditions: // 1. Node's left's right pointer should be // current node. // 2. Node's right's left pointer should be // current node. // If one condition is met it is guaranteed // that the other condition is also true. return node->left && node->left->right == node && node->right && node->right->left == node; } // Compute the height of a tree int findTreeHeight(struct Node* node) { // if node is NULL return -1. if (node == NULL) return -1; // if node is a leaf node return 0 if (isLeaf(node)) return 0; // compute the depth of each subtree and take maximum int leftDepth = findTreeHeight(node->left); int rightDepth = findTreeHeight(node->right); return 1 + (leftDepth > rightDepth ? leftDepth : rightDepth); } struct Node* createNode(int data) { struct Node* newNode = (struct Node*)malloc(sizeof(struct Node)); newNode->data = data; newNode->left = NULL; newNode->right = NULL; return newNode; } int main() { struct Node* root = createNode(1); root->left = createNode(2); root->right = createNode(3); root->left->left = createNode(4); root->left->right = createNode(5); root->left->left->left = createNode(6); // Given tree contains 3 leaf nodes struct Node* l1 = root->left->left->left; struct Node* l2 = root->left->right; struct Node* l3 = root->right; // create circular doubly linked list out of // leaf nodes of the tree // set next pointer of linked list l1->right = l2 l2->right = l3 l3->right = l1; // set prev pointer of linked list l3->left = l2 l2->left = l1 l1->left = l3; printf('%d' findTreeHeight(root)); return 0; }
Java // Java program to calculate height of a special tree // whose leaf nodes forms a circular doubly linked list class Node { int data; Node left right; Node(int x) { data = x; left = null; right = null; } } class GfG { // function to check if given // node is a leaf node or node static boolean isLeaf(Node node) { // For a node to be a leaf node it should // satisfy the following two conditions: // 1. Node's left's right pointer should be // current node. // 2. Node's right's left pointer should be // current node. // If one condition is met it is guaranteed // that the other condition is also true. return node.left != null && node.left.right == node && node.right != null && node.right.left == node; } // Compute the height of a tree static int findTreeHeight(Node node) { // if node is NULL return -1. if (node == null) return -1; // if node is a leaf node return 0 if (isLeaf(node)) return 0; // compute the depth of each subtree and take maximum return 1 + Math.max(findTreeHeight(node.left) findTreeHeight(node.right)); } public static void main(String[] args) { Node root = new Node(1); root.left = new Node(2); root.right = new Node(3); root.left.left = new Node(4); root.left.right = new Node(5); root.left.left.left = new Node(6); // Given tree contains 3 leaf nodes Node l1 = root.left.left.left; Node l2 = root.left.right; Node l3 = root.right; // create circular doubly linked list out of // leaf nodes of the tree // set next pointer of linked list l1.right = l2; l2.right = l3; l3.right = l1; // set prev pointer of linked list l3.left = l2; l2.left = l1; l1.left = l3; System.out.println(findTreeHeight(root)); } }
Python # Python program to calculate height of a special tree # whose leaf nodes forms a circular doubly linked list class Node: def __init__(self data): self.data = data self.left = None self.right = None # function to check if given # node is a leaf node or node def isLeaf(node): # For a node to be a leaf node it should # satisfy the following two conditions: # 1. Node's left's right pointer should be # current node. # 2. Node's right's left pointer should be # current node. # If one condition is met it is guaranteed # that the other condition is also true. return (node.left and node.left.right == node and node.right and node.right.left == node) # Compute the height of a tree def findTreeHeight(node): # if node is NULL return -1. if node is None: return -1 # if node is a leaf node return 0 if isLeaf(node): return 0 # compute the depth of each subtree and take maximum return 1 + max(findTreeHeight(node.left) findTreeHeight(node.right)) if __name__ == '__main__': root = Node(1) root.left = Node(2) root.right = Node(3) root.left.left = Node(4) root.left.right = Node(5) root.left.left.left = Node(6) # Given tree contains 3 leaf nodes l1 = root.left.left.left l2 = root.left.right l3 = root.right # create circular doubly linked list out of # leaf nodes of the tree # set next pointer of linked list l1.right = l2 l2.right = l3 l3.right = l1 # set prev pointer of linked list l3.left = l2 l2.left = l1 l1.left = l3 print(findTreeHeight(root))
C# // C# program to calculate height of a special tree // whose leaf nodes forms a circular doubly linked list using System; class Node { public int data; public Node left right; public Node(int x) { data = x; left = null; right = null; } } class GfG { // function to check if given // node is a leaf node or node static bool isLeaf(Node node) { // For a node to be a leaf node it should // satisfy the following two conditions: // 1. Node's left's right pointer should be // current node. // 2. Node's right's left pointer should be // current node. // If one condition is met it is guaranteed // that the other condition is also true. return node.left != null && node.left.right == node && node.right != null && node.right.left == node; } // Compute the height of a tree static int findTreeHeight(Node node) { // if node is NULL return -1. if (node == null) return -1; // if node is a leaf node return 0 if (isLeaf(node)) return 0; // compute the depth of each subtree and take maximum return 1 + Math.Max(findTreeHeight(node.left) findTreeHeight(node.right)); } static void Main(string[] args) { Node root = new Node(1); root.left = new Node(2); root.right = new Node(3); root.left.left = new Node(4); root.left.right = new Node(5); root.left.left.left = new Node(6); // Given tree contains 3 leaf nodes Node l1 = root.left.left.left; Node l2 = root.left.right; Node l3 = root.right; // create circular doubly linked list out of // leaf nodes of the tree // set next pointer of linked list l1.right = l2; l2.right = l3; l3.right = l1; // set prev pointer of linked list l3.left = l2; l2.left = l1; l1.left = l3; Console.WriteLine(findTreeHeight(root)); } }
JavaScript // JavaScript program to calculate height of a special tree // whose leaf nodes forms a circular doubly linked list class Node { constructor(data) { this.data = data; this.left = null; this.right = null; } } // function to check if given // node is a leaf node or node function isLeaf(node) { // For a node to be a leaf node it should // satisfy the following two conditions: // 1. Node's left's right pointer should be // current node. // 2. Node's right's left pointer should be // current node. // If one condition is met it is guaranteed // that the other condition is also true. return node.left && node.left.right === node && node.right && node.right.left === node; } // Compute the height of a tree function findTreeHeight(node) { // if node is NULL return -1. if (node === null) return -1; // if node is a leaf node return 0 if (isLeaf(node)) return 0; // compute the depth of each subtree and take maximum return 1 + Math.max(findTreeHeight(node.left) findTreeHeight(node.right)); } const root = new Node(1); root.left = new Node(2); root.right = new Node(3); root.left.left = new Node(4); root.left.right = new Node(5); root.left.left.left = new Node(6); // Given tree contains 3 leaf nodes const l1 = root.left.left.left; const l2 = root.left.right; const l3 = root.right; // create circular doubly linked list out of // leaf nodes of the tree // set next pointer of linked list l1.right = l2; l2.right = l3; l3.right = l1; // set prev pointer of linked list l3.left = l2; l2.left = l1; l1.left = l3; console.log(findTreeHeight(root));
Çıkış
3
Zaman Karmaşıklığı: O(n) nerede N düğüm sayısıdır.
Yardımcı alan: Ah)