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מצא גובה של עץ בינארי מיוחד שצמתי העלים שלו מחוברים

בהינתן א עץ בינארי מיוחד שֶׁל מִי צמתים עלים מחוברים לצורה א רשימה מעגלית מקושרת כפולה המשימה היא למצוא את גוֹבַה של העץ.

דוגמאות:



קֶלֶט:

רִבּוּב
מצא-גובה-של-עץ-בינארי-מיוחד-שצמתי העלים-מחוברים-2' title=

תְפוּקָה: 2
הֶסבֵּר: גובה העץ הבינארי לאחר זיהוי צמתי העלים הוא 2. בעץ הבינארי לעיל 6 5 ו-4 הם צמתים עלים והם יוצרים רשימה מעגלית מקושרת כפולה. כאן המצביע השמאלי של צומת העלה יפעל כמצביע קודם של רשימה עגולה מקושרת כפולה והמצביע הימני שלו יפעל כמצביע הבא של רשימה מעגלית מקושרת כפולה. 

קֶלֶט:



מצא-גובה-של-עץ-בינארי-מיוחד שצמתי העלים-מחוברים-1' loading='lazy' title=

תְפוּקָה: 1
הֶסבֵּר: גובה העץ הבינארי לאחר זיהוי צמתי העלים הוא 1. בעץ הבינארי 2 ו-3 לעיל הם צמתים עלים והם יוצרים רשימה מעגלית מקושרת כפולה.

גִישָׁה :

הרעיון הוא לעקוב גישה דומה כפי שאנו עושים עבור מציאת גובה של עץ בינארי רגיל . אָנוּ באופן רקורסיבי לְחַשֵׁב גוֹבַה שֶׁל שמאל וימין תת-עצים של צומת והקצאה גוֹבַה לצומת כ מקסימום מהגובה של שני ילדים פלוס 1. אבל ילד שמאל וימין של א צומת עלה הם אפס עבור עצים בינאריים רגילים. אבל כאן צומת עלים הוא צומת רשימה מעגלית מקושרת כפולה. אז כדי שצומת יהיה צומת עלה אנחנו בודקים אם הצומת משמאל לימין מצביע על צוֹמֶת ואת שלה ימין לשמאל מצביע גם על צוֹמֶת עצמו.



C++
// 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)); 

תְפוּקָה
3

מורכבות זמן: O(n) איפה נ הוא מספר הצמתים.
חלל עזר: O(h)