给定一个单链表，其数据成员按升序排序。构建一个 平衡二叉搜索树 其数据成员与给定的链表相同。 例如：
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Input: Linked List 1->2->3Output: A Balanced BST 2 / 1 3 Input: Linked List 1->2->3->4->5->6->7Output: A Balanced BST 4 / 2 6 / / 1 3 5 7 Input: Linked List 1->2->3->4Output: A Balanced BST 3 / 2 4 / 1Input: Linked List 1->2->3->4->5->6Output: A Balanced BST 4 / 2 6 / / 1 3 5
方法1（简单） 下面是一个简单的算法，我们首先找到列表的中间节点，并将其作为要构造的树的根。
1) Get the Middle of the linked list and make it root.2) Recursively do same for the left half and right half. a) Get the middle of the left half and make it left child of the root created in step 1. b) Get the middle of right half and make it the right child of the root created in step 1.
时间复杂度：O（nLogn），其中n是链表中的节点数。 方法2（棘手） 方法1从根到叶构建树。在这种方法中，我们从叶子到根进行构造。其思想是在BST中插入节点的顺序与它们在链表中出现的顺序相同，这样就可以以O（n）时间复杂度构建树。我们首先计算给定链表中的节点数。设计数为n。在计算节点数之后，我们取左n/2个节点，递归地构造左子树。构造左子树后，我们为根分配内存，并将左子树与根连接起来。最后，我们递归地构造正确的子树并将其与根连接起来。 在构造BST的同时，我们还将列表头指针移动到下一个，以便在每个递归调用中都有适当的指针。
下面是方法2的实现。创建平衡BST的主代码突出显示。
C++
// C++ implementation of above approach #include <bits/stdc++.h> using namespace std; /* Link list node */ class LNode { public : int data; LNode* next; }; /* A Binary Tree node */ class TNode { public : int data; TNode* left; TNode* right; }; TNode* newNode( int data); int countLNodes(LNode *head); TNode* sortedListToBSTRecur(LNode **head_ref, int n); /* This function counts the number of nodes in Linked List and then calls sortedListToBSTRecur() to construct BST */ TNode* sortedListToBST(LNode *head) { /*Count the number of nodes in Linked List */ int n = countLNodes(head); /* Construct BST */ return sortedListToBSTRecur(&head, n); } /* The main function that constructs balanced BST and returns root of it. head_ref --> Pointer to pointer to head node of linked list n --> No. of nodes in Linked List */ TNode* sortedListToBSTRecur(LNode **head_ref, int n) { /* Base Case */ if (n <= 0) return NULL; /* Recursively construct the left subtree */ TNode *left = sortedListToBSTRecur(head_ref, n/2); /* Allocate memory for root, and link the above constructed left subtree with root */ TNode *root = newNode((*head_ref)->data); root->left = left; /* Change head pointer of Linked List for parent recursive calls */ *head_ref = (*head_ref)->next; /* Recursively construct the right subtree and link it with root The number of nodes in right subtree is total nodes - nodes in left subtree - 1 (for root) which is n-n/2-1*/ root->right = sortedListToBSTRecur(head_ref, n - n / 2 - 1); return root; } /* UTILITY FUNCTIONS */ /* A utility function that returns count of nodes in a given Linked List */ int countLNodes(LNode *head) { int count = 0; LNode *temp = head; while (temp) { temp = temp->next; count++; } return count; } /* Function to insert a node at the beginning of the linked list */ void push(LNode** head_ref, int new_data) { /* allocate node */ LNode* new_node = new LNode(); /* put in the data */ new_node->data = new_data; /* link the old list off the new node */ new_node->next = (*head_ref); /* move the head to point to the new node */ (*head_ref) = new_node; } /* Function to print nodes in a given linked list */ void printList(LNode *node) { while (node!=NULL) { cout << node->data << " " ; node = node->next; } } /* Helper function that allocates a new node with the given data and NULL left and right pointers. */ TNode* newNode( int data) { TNode* node = new TNode(); node->data = data; node->left = NULL; node->right = NULL; return node; } /* A utility function to print preorder traversal of BST */ void preOrder(TNode* node) { if (node == NULL) return ; cout<<node->data<< " " ; preOrder(node->left); preOrder(node->right); } /* Driver code*/ int main() { /* Start with the empty list */ LNode* head = NULL; /* Let us create a sorted linked list to test the functions Created linked list will be 1->2->3->4->5->6->7 */ push(&head, 7); push(&head, 6); push(&head, 5); push(&head, 4); push(&head, 3); push(&head, 2); push(&head, 1); cout<< "Given Linked List " ; printList(head); /* Convert List to BST */ TNode *root = sortedListToBST(head); cout<< "PreOrder Traversal of constructed BST " ; preOrder(root); return 0; } // This code is contributed by rathbhupendra |
C
#include<stdio.h> #include<stdlib.h> /* Link list node */ struct LNode { int data; struct LNode* next; }; /* A Binary Tree node */ struct TNode { int data; struct TNode* left; struct TNode* right; }; struct TNode* newNode( int data); int countLNodes( struct LNode *head); struct TNode* sortedListToBSTRecur( struct LNode **head_ref, int n); /* This function counts the number of nodes in Linked List and then calls sortedListToBSTRecur() to construct BST */ struct TNode* sortedListToBST( struct LNode *head) { /*Count the number of nodes in Linked List */ int n = countLNodes(head); /* Construct BST */ return sortedListToBSTRecur(&head, n); } /* The main function that constructs balanced BST and returns root of it. head_ref --> Pointer to pointer to head node of linked list n --> No. of nodes in Linked List */ struct TNode* sortedListToBSTRecur( struct LNode **head_ref, int n) { /* Base Case */ if (n <= 0) return NULL; /* Recursively construct the left subtree */ struct TNode *left = sortedListToBSTRecur(head_ref, n/2); /* Allocate memory for root, and link the above constructed left subtree with root */ struct TNode *root = newNode((*head_ref)->data); root->left = left; /* Change head pointer of Linked List for parent recursive calls */ *head_ref = (*head_ref)->next; /* Recursively construct the right subtree and link it with root The number of nodes in right subtree is total nodes - nodes in left subtree - 1 (for root) which is n-n/2-1*/ root->right = sortedListToBSTRecur(head_ref, n-n/2-1); return root; } /* UTILITY FUNCTIONS */ /* A utility function that returns count of nodes in a given Linked List */ int countLNodes( struct LNode *head) { int count = 0; struct LNode *temp = head; while (temp) { temp = temp->next; count++; } return count; } /* Function to insert a node at the beginning of the linked list */ void push( struct LNode** head_ref, int new_data) { /* allocate node */ struct LNode* new_node = ( struct LNode*) malloc ( sizeof ( struct LNode)); /* put in the data */ new_node->data = new_data; /* link the old list off the new node */ new_node->next = (*head_ref); /* move the head to point to the new node */ (*head_ref) = new_node; } /* Function to print nodes in a given linked list */ void printList( struct LNode *node) { while (node!=NULL) { printf ( "%d " , node->data); node = node->next; } } /* Helper function that allocates a new node with the given data and NULL left and right pointers. */ struct TNode* newNode( int data) { struct TNode* node = ( struct TNode*) malloc ( sizeof ( struct TNode)); node->data = data; node->left = NULL; node->right = NULL; return node; } /* A utility function to print preorder traversal of BST */ void preOrder( struct TNode* node) { if (node == NULL) return ; printf ( "%d " , node->data); preOrder(node->left); preOrder(node->right); } /* Driver program to test above functions*/ int main() { /* Start with the empty list */ struct LNode* head = NULL; /* Let us create a sorted linked list to test the functions Created linked list will be 1->2->3->4->5->6->7 */ push(&head, 7); push(&head, 6); push(&head, 5); push(&head, 4); push(&head, 3); push(&head, 2); push(&head, 1); printf ( " Given Linked List " ); printList(head); /* Convert List to BST */ struct TNode *root = sortedListToBST(head); printf ( " PreOrder Traversal of constructed BST " ); preOrder(root); return 0; } |
JAVA
class LinkedList { /* head node of link list */ static LNode head; /* Link list Node */ class LNode { int data; LNode next, prev; LNode( int d) { data = d; next = prev = null ; } } /* A Binary Tree Node */ class TNode { int data; TNode left, right; TNode( int d) { data = d; left = right = null ; } } /* This function counts the number of nodes in Linked List and then calls sortedListToBSTRecur() to construct BST */ TNode sortedListToBST() { /*Count the number of nodes in Linked List */ int n = countNodes(head); /* Construct BST */ return sortedListToBSTRecur(n); } /* The main function that constructs balanced BST and returns root of it. n --> No. of nodes in the Doubly Linked List */ TNode sortedListToBSTRecur( int n) { /* Base Case */ if (n <= 0 ) return null ; /* Recursively construct the left subtree */ TNode left = sortedListToBSTRecur(n / 2 ); /* head_ref now refers to middle node, make middle node as root of BST*/ TNode root = new TNode(head.data); // Set pointer to left subtree root.left = left; /* Change head pointer of Linked List for parent recursive calls */ head = head.next; /* Recursively construct the right subtree and link it with root. The number of nodes in right subtree is total nodes - nodes in left subtree - 1 (for root) */ root.right = sortedListToBSTRecur(n - n / 2 - 1 ); return root; } /* UTILITY FUNCTIONS */ /* A utility function that returns count of nodes in a given Linked List */ int countNodes(LNode head) { int count = 0 ; LNode temp = head; while (temp != null ) { temp = temp.next; count++; } return count; } /* Function to insert a node at the beginning of the Doubly Linked List */ void push( int new_data) { /* allocate node */ LNode new_node = new LNode(new_data); /* since we are adding at the beginning, prev is always NULL */ new_node.prev = null ; /* link the old list off the new node */ new_node.next = head; /* change prev of head node to new node */ if (head != null ) head.prev = new_node; /* move the head to point to the new node */ head = new_node; } /* Function to print nodes in a given linked list */ void printList(LNode node) { while (node != null ) { System.out.print(node.data + " " ); node = node.next; } } /* A utility function to print preorder traversal of BST */ void preOrder(TNode node) { if (node == null ) return ; System.out.print(node.data + " " ); preOrder(node.left); preOrder(node.right); } /* Driver program to test above functions */ public static void main(String[] args) { LinkedList llist = new LinkedList(); /* Let us create a sorted linked list to test the functions Created linked list will be 7->6->5->4->3->2->1 */ llist.push( 7 ); llist.push( 6 ); llist.push( 5 ); llist.push( 4 ); llist.push( 3 ); llist.push( 2 ); llist.push( 1 ); System.out.println( "Given Linked List " ); llist.printList(head); /* Convert List to BST */ TNode root = llist.sortedListToBST(); System.out.println( "" ); System.out.println( "Pre-Order Traversal of constructed BST " ); llist.preOrder(root); } } // This code has been contributed by Mayank Jaiswal(mayank_24) |
Python3
# Python3 implementation of above approach # Link list node class LNode : def __init__( self ): self .data = None self . next = None # A Binary Tree node class TNode : def __init__( self ): self .data = None self .left = None self .right = None head = None # This function counts the number of # nodes in Linked List and then calls # sortedListToBSTRecur() to construct BST def sortedListToBST(): global head # Count the number of nodes in Linked List n = countLNodes(head) # Construct BST return sortedListToBSTRecur(n) # The main function that constructs # balanced BST and returns root of it. # head -. Pointer to pointer to # head node of linked list n -. No. # of nodes in Linked List def sortedListToBSTRecur( n) : global head # Base Case if (n < = 0 ) : return None # Recursively construct the left subtree left = sortedListToBSTRecur( int (n / 2 )) # Allocate memory for root, and # link the above constructed left # subtree with root root = newNode((head).data) root.left = left # Change head pointer of Linked List # for parent recursive calls head = (head). next # Recursively construct the right # subtree and link it with root # The number of nodes in right subtree # is total nodes - nodes in # left subtree - 1 (for root) which is n-n/2-1 root.right = sortedListToBSTRecur( n - int (n / 2 ) - 1 ) return root # UTILITY FUNCTIONS # A utility function that returns # count of nodes in a given Linked List def countLNodes(head) : count = 0 temp = head while (temp ! = None ): temp = temp. next count = count + 1 return count # Function to insert a node #at the beginning of the linked list def push(head, new_data) : # allocate node new_node = LNode() # put in the data new_node.data = new_data # link the old list off the new node new_node. next = (head) # move the head to point to the new node (head) = new_node return head # Function to print nodes in a given linked list def printList(node): while (node ! = None ): print ( node.data ,end = " " ) node = node. next # Helper function that allocates a new node with the # given data and None left and right pointers. def newNode(data) : node = TNode() node.data = data node.left = None node.right = None return node # A utility function to # print preorder traversal of BST def preOrder( node) : if (node = = None ) : return print (node.data, end = " " ) preOrder(node.left) preOrder(node.right) # Driver code # Start with the empty list head = None # Let us create a sorted linked list to test the functions # Created linked list will be 1.2.3.4.5.6.7 head = push(head, 7 ) head = push(head, 6 ) head = push(head, 5 ) head = push(head, 4 ) head = push(head, 3 ) head = push(head, 2 ) head = push(head, 1 ) print ( "Given Linked List " ) printList(head) # Convert List to BST root = sortedListToBST() print ( "PreOrder Traversal of constructed BST " ) preOrder(root) # This code is contributed by Arnab Kundu |
C#
// C# implementation of above approach using System; public class LinkedList { /* head node of link list */ static LNode head; /* Link list Node */ class LNode { public int data; public LNode next, prev; public LNode( int d) { data = d; next = prev = null ; } } /* A Binary Tree Node */ class TNode { public int data; public TNode left, right; public TNode( int d) { data = d; left = right = null ; } } /* This function counts the number of nodes in Linked List and then calls sortedListToBSTRecur() to construct BST */ TNode sortedListToBST() { /*Count the number of nodes in Linked List */ int n = countNodes(head); /* Construct BST */ return sortedListToBSTRecur(n); } /* The main function that constructs balanced BST and returns root of it. n --> No. of nodes in the Doubly Linked List */ TNode sortedListToBSTRecur( int n) { /* Base Case */ if (n <= 0) return null ; /* Recursively construct the left subtree */ TNode left = sortedListToBSTRecur(n / 2); /* head_ref now refers to middle node, make middle node as root of BST*/ TNode root = new TNode(head.data); // Set pointer to left subtree root.left = left; /* Change head pointer of Linked List for parent recursive calls */ head = head.next; /* Recursively construct the right subtree and link it with root. The number of nodes in right subtree is total nodes - nodes in left subtree - 1 (for root) */ root.right = sortedListToBSTRecur(n - n / 2 - 1); return root; } /* UTILITY FUNCTIONS */ /* A utility function that returns count of nodes in a given Linked List */ int countNodes(LNode head) { int count = 0; LNode temp = head; while (temp != null ) { temp = temp.next; count++; } return count; } /* Function to insert a node at the beginning of the Doubly Linked List */ void push( int new_data) { /* allocate node */ LNode new_node = new LNode(new_data); /* since we are adding at the beginning, prev is always NULL */ new_node.prev = null ; /* link the old list off the new node */ new_node.next = head; /* change prev of head node to new node */ if (head != null ) head.prev = new_node; /* move the head to point to the new node */ head = new_node; } /* Function to print nodes in a given linked list */ void printList(LNode node) { while (node != null ) { Console.Write(node.data + " " ); node = node.next; } } /* A utility function to print preorder traversal of BST */ void preOrder(TNode node) { if (node == null ) return ; Console.Write(node.data + " " ); preOrder(node.left); preOrder(node.right); } /* Driver code */ public static void Main(String[] args) { LinkedList llist = new LinkedList(); /* Let us create a sorted linked list to test the functions Created linked list will be 7->6->5->4->3->2->1 */ llist.push(7); llist.push(6); llist.push(5); llist.push(4); llist.push(3); llist.push(2); llist.push(1); Console.WriteLine( "Given Linked List " ); llist.printList(head); /* Convert List to BST */ TNode root = llist.sortedListToBST(); Console.WriteLine( "" ); Console.WriteLine( "Pre-Order Traversal of constructed BST " ); llist.preOrder(root); } } // This code is contributed by Rajput-Ji |
Javascript
<script> // JavaScript implementation of above approach /* head node of link list */ var head = null ; /* Link list Node */ class LNode { constructor(d) { this .data = d; this .next = null ; this .prev = null ; } } /* A Binary Tree Node */ class TNode { constructor(d) { this .data = d; this .left = null ; this .right = null ; } } /* This function counts the number of nodes in Linked List and then calls sortedListToBSTRecur() to construct BST */ function sortedListToBST() { /*Count the number of nodes in Linked List */ var n = countNodes(head); /* Construct BST */ return sortedListToBSTRecur(n); } /* The main function that constructs balanced BST and returns root of it. n --> No. of nodes in the Doubly Linked List */ function sortedListToBSTRecur(n) { /* Base Case */ if (n <= 0) return null ; /* Recursively construct the left subtree */ var left = sortedListToBSTRecur(parseInt(n / 2)); /* head_ref now refers to middle node, make middle node as root of BST*/ var root = new TNode(head.data); // Set pointer to left subtree root.left = left; /* Change head pointer of Linked List for parent recursive calls */ head = head.next; /* Recursively construct the right subtree and link it with root. The number of nodes in right subtree is total nodes - nodes in left subtree - 1 (for root) */ root.right = sortedListToBSTRecur(n - parseInt(n / 2) - 1); return root; } /* UTILITY FUNCTIONS */ /* A utility function that returns count of nodes in a given Linked List */ function countNodes(head) { var count = 0; var temp = head; while (temp != null ) { temp = temp.next; count++; } return count; } /* Function to insert a node at the beginning of the Doubly Linked List */ function push( new_data) { /* allocate node */ var new_node = new LNode(new_data); /* since we are adding at the beginning, prev is always NULL */ new_node.prev = null ; /* link the old list off the new node */ new_node.next = head; /* change prev of head node to new node */ if (head != null ) head.prev = new_node; /* move the head to point to the new node */ head = new_node; } /* Function to print nodes in a given linked list */ function printList( node) { while (node != null ) { document.write(node.data + " " ); node = node.next; } } /* A utility function to print preorder traversal of BST */ function preOrder(node) { if (node == null ) return ; document.write(node.data + " " ); preOrder(node.left); preOrder(node.right); } /* Driver code */ /* Let us create a sorted linked list to test the functions Created linked list will be 7->6->5->4->3->2->1 */ push(7); push(6); push(5); push(4); push(3); push(2); push(1); document.write( "Given Linked List " ); printList(head); /* Convert List to BST */ var root = sortedListToBST(); document.write( "<br>" ); document.write( "Pre-Order Traversal of constructed BST " ); preOrder(root); </script> |
输出：
Given Linked List 1 2 3 4 5 6 7 PreOrder Traversal of constructed BST 4 2 1 3 6 5 7
时间复杂性： O（n） 如果您发现任何不正确的地方，或者您想分享有关上述主题的更多信息，请写下评论。
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