Sorted DLL 到 Balanced BST 的就地转换

给定一个双向链表，其中的数据成员按升序排列。构造一个平衡二叉搜索树，它与给定的双向链表具有相同的数据成员。必须就地构造树（不应为树转换分配新节点）
例子：

Input:  Doubly Linked List 1  2  3
Output: A Balanced BST 
     2   
   /  \  
  1    3 


Input: Doubly Linked List 1  2 3  4 5  6  7
Output: A Balanced BST
        4
      /   \
     2     6
   /  \   / \
  1   3  5   7  

Input: Doubly Linked List 1  2  3  4
Output: A Balanced BST
      3   
    /  \  
   2    4 
 / 
1

Input:  Doubly Linked List 1  2  3  4  5  6
Output: A Balanced BST
      4   
    /   \  
   2     6 
 /  \   / 
1   3  5   

双链表转换与这个单链表问题非常相似，方法 1 与前一篇文章的方法 1 完全相同。方法2也差不多。方法 2 的唯一区别是，我们重用相同的 DLL 节点，而不是为 BST 分配新节点。我们使用上一个指针作为左指针，下一个指针作为右指针。
方法一（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，并使其成为要构造的树的根。

1) Get the Middle of the linked list and make it root.
2) Recursively do same for left half and right half.
       a) Get the middle of left half and make it left child of the root
          created in step 1.
       b) Get the middle of right half and make it 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。
方法二（棘手）
方法1从根到叶构造树。在这种方法中，我们从叶子到根构造。其思想是按照与双向链表中出现的顺序相同的顺序在 BST 中插入节点，以便可以在 O(n) 时间复杂度内构建树。我们首先计算给定链表中的节点数。让计数为 n。在计算节点数后，我们取左 n/2 个节点并递归构建左子树。构建左子树后，我们将中间节点分配给根并将左子树与根链接。最后，我们递归地构造右子树并将其与根链接。
在构建 BST 的同时，我们还不断将列表头指针移动到 next 以便我们在每次递归调用中都有适当的指针。
下面是方法2的实现。高亮显示了创建Balanced BST的主要代码。

C++

#include 
using namespace std;
 
/* A Doubly Linked List node that
will also be used as a tree node */
class Node
{
    public:
    int data;
 
    // For tree, next pointer can be
    // used as right subtree pointer
    Node* next;
 
    // For tree, prev pointer can be
    // used as left subtree pointer
    Node* prev;
};
 
// A utility function to count nodes in a Linked List
int countNodes(Node *head);
 
Node* sortedListToBSTRecur(Node **head_ref, int n);
 
/* This function counts the number of
nodes in Linked List and then calls
sortedListToBSTRecur() to construct BST */
Node* sortedListToBST(Node *head)
{
    /*Count the number of nodes in Linked List */
    int n = countNodes(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 Doubly linked list
n --> No. of nodes in the Doubly Linked List */
Node* sortedListToBSTRecur(Node **head_ref, int n)
{
    /* Base Case */
    if (n <= 0)
        return NULL;
 
    /* Recursively construct the left subtree */
    Node *left = sortedListToBSTRecur(head_ref, n/2);
 
    /* head_ref now refers to middle node,
    make middle node as root of BST*/
    Node *root = *head_ref;
 
    // Set pointer to left subtree
    root->prev = 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) */
    root->next = 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 countNodes(Node *head)
{
    int count = 0;
    Node *temp = head;
    while(temp)
    {
        temp = temp->next;
        count++;
    }
    return count;
}
 
/* Function to insert a node at
the beginning of the Doubly Linked List */
void push(Node** head_ref, int new_data)
{
    /* allocate node */
    Node* new_node = new Node();
 
    /* put in the data */
    new_node->data = 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_ref);
 
    /* change prev of head node to new node */
    if((*head_ref) != NULL)
    (*head_ref)->prev = new_node ;
 
    /* move the head to point to the new node */
    (*head_ref) = new_node;
}
 
/* Function to print nodes in a given linked list */
void printList(Node *node)
{
    while (node!=NULL)
    {
        cout<data<<" ";
        node = node->next;
    }
}
 
/* A utility function to print
preorder traversal of BST */
void preOrder(Node* node)
{
    if (node == NULL)
        return;
    cout<data<<" ";
    preOrder(node->prev);
    preOrder(node->next);
}
 
/* Driver code*/
int main()
{
    /* Start with the empty list */
    Node* head = NULL;
 
    /* Let us create a sorted linked list to test the functions
    Created linked list will be 7->6->5->4->3->2->1 */
    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\n";
    printList(head);
 
    /* Convert List to BST */
    Node *root = sortedListToBST(head);
    cout<<"\nPreOrder Traversal of constructed BST \n ";
    preOrder(root);
 
    return 0;
}
 
// This code is contributed by rathbhupendra

C

#include
#include
 
/* A Doubly Linked List node that will also be used as a tree node */
struct Node
{
    int data;
 
    // For tree, next pointer can be used as right subtree pointer
    struct Node* next;
 
    // For tree, prev pointer can be used as left subtree pointer
    struct Node* prev;
};
 
// A utility function to count nodes in a Linked List
int countNodes(struct Node *head);
 
struct Node* sortedListToBSTRecur(struct Node **head_ref, int n);
 
/* This function counts the number of nodes in Linked List and then calls
   sortedListToBSTRecur() to construct BST */
struct Node* sortedListToBST(struct Node *head)
{
    /*Count the number of nodes in Linked List */
    int n = countNodes(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 Doubly linked list
       n  --> No. of nodes in the Doubly Linked List */
struct Node* sortedListToBSTRecur(struct Node **head_ref, int n)
{
    /* Base Case */
    if (n <= 0)
        return NULL;
 
    /* Recursively construct the left subtree */
    struct Node *left = sortedListToBSTRecur(head_ref, n/2);
 
    /* head_ref now refers to middle node, make middle node as root of BST*/
    struct Node *root = *head_ref;
 
    // Set pointer to left subtree
    root->prev = 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) */
    root->next = 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 countNodes(struct Node *head)
{
    int count = 0;
    struct Node *temp = head;
    while(temp)
    {
        temp = temp->next;
        count++;
    }
    return count;
}
 
/* Function to insert a node at the beginning of the Doubly Linked List */
void push(struct Node** head_ref, int new_data)
{
    /* allocate node */
    struct Node* new_node =
            (struct Node*) malloc(sizeof(struct Node));
 
    /* put in the data  */
    new_node->data  = 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_ref);
 
    /* change prev of head node to new node */
    if((*head_ref) !=  NULL)
      (*head_ref)->prev = new_node ;
 
    /* 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 Node *node)
{
    while (node!=NULL)
    {
        printf("%d ", node->data);
        node = node->next;
    }
}
 
/* A utility function to print preorder traversal of BST */
void preOrder(struct Node* node)
{
    if (node == NULL)
        return;
    printf("%d ", node->data);
    preOrder(node->prev);
    preOrder(node->next);
}
 
/* Driver program to test above functions*/
int main()
{
    /* Start with the empty list */
    struct Node* head = NULL;
 
    /* Let us create a sorted linked list to test the functions
     Created linked list will be 7->6->5->4->3->2->1 */
    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\n");
    printList(head);
 
    /* Convert List to BST */
    struct Node *root = sortedListToBST(head);
    printf("\n PreOrder Traversal of constructed BST \n ");
    preOrder(root);
 
    return 0;
}

Java

class Node
{
    int data;
    Node next, prev;
 
    Node(int d)
    {
        data = d;
        next = prev = null;
    }
}
 
class LinkedList
{
    Node head;
 
    /* This function counts the number of nodes in Linked List
       and then calls sortedListToBSTRecur() to construct BST */
    Node 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 */
    Node sortedListToBSTRecur(int n)
    {
        /* Base Case */
        if (n <= 0)
            return null;
 
        /* Recursively construct the left subtree */
        Node left = sortedListToBSTRecur(n / 2);
 
        /* head_ref now refers to middle node,
           make middle node as root of BST*/
        Node root = head;
 
        // Set pointer to left subtree
        root.prev = 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.next = sortedListToBSTRecur(n - n / 2 - 1);
 
        return root;
    }
 
    /* UTILITY FUNCTIONS */
    /* A utility function that returns count of nodes in a
       given Linked List */
    int countNodes(Node head)
    {
        int count = 0;
        Node 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 */
        Node new_node = new Node(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()
    {
        Node node = head;
        while (node != null)
        {
            System.out.print(node.data + " ");
            node = node.next;
        }
    }
 
    /* A utility function to print preorder traversal of BST */
    void preOrder(Node node)
    {
        if (node == null)
            return;
        System.out.print(node.data + " ");
        preOrder(node.prev);
        preOrder(node.next);
    }
 
    /* Drier 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();
 
        /* Convert List to BST */
        Node 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)

Javascript

 
class Node
{
    constructor(d)
    {
        this.data=d;
        this.next=this.prev=null;
    }
}
 
let head;
 
/* 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 */
        let 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 */
        let left = sortedListToBSTRecur(Math.floor(n / 2));
  
        /* head_ref now refers to middle node,
           make middle node as root of BST*/
        let root = head;
  
        // Set pointer to left subtree
        root.prev = 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.next = sortedListToBSTRecur(n - Math.floor(n / 2) - 1);
  
        return root;
}
 
/* UTILITY FUNCTIONS */
    /* A utility function that returns count of nodes in a
       given Linked List */
function countNodes(head)
{
    let count = 0;
        let 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 */
        let new_node = new Node(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()
{
    let node = head;
        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.prev);
        preOrder(node.next);
}
 
/* Drier program to test above functions */
/* 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 <br>");
printList();
 
/* Convert List to BST */
let root = sortedListToBST();
document.write("<br>");
document.write("Pre-Order Traversal of constructed BST <br>");
preOrder(root);
     
// This code is contributed by avanitrachhadiya2155

输出：

Given Linked List 
1 2 3 4 5 6 7 
Pre-Order Traversal of constructed BST 
4 2 1 3 6 5 7 

如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。