已排序 DLL 到平衡 BST 的原地转换
原文：https://www.geeksforgeeks.org/in-place-conversion-of-sorted-dll-to-balanced-bst/
给定一个双链表，该列表具有按升序排序的数据成员。 构造一个平衡二叉搜索树，该树的数据成员与给定的双链表相同。 该树必须原地构建（不应为树转换分配新节点）
示例：
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 4 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
双链表转换非常类似于这个单链表问题，并且方法 1 与先前文章的方法 1 完全相同。 方法 2 也几乎相同。 方法 2 的唯一区别是，我们没有重复为 BST 分配新节点，而是重用了相同的 DLL 节点。 我们将上一个指针用作左侧，将下一个指针用作右侧。
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，并将其设为要构建的树的根。
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 2（困难）
方法 1 构造了从根到叶的树。 在这种方法中，我们从叶到根进行构建。 想法是按照与出现在双链表中的顺序相同的顺序在 BST 中插入节点，以便可以以O(n)时间复杂度构造树。 我们首先计算给定链表中的节点数。 让计数为n。 在计数节点之后，我们取左n / 2个节点并递归构造左子树。 构造左子树后，我们将中间节点分配给根，并将左子树与根链接。 最后，我们递归构造正确的子树并将其与根链接。
在构造 BST 时，我们还不断将列表头指针移到next位置，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。突出显示了创建平衡 BST 的主要代码。
C++
#include <bits/stdc++.h>
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 beginging 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<<node->data<<" ";
node = node->next;
}
}
/* A utility function to print
preorder traversal of BST */
void preOrder(Node* node)
{
if (node == NULL)
return;
cout<<node->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<stdio.h>
#include<stdlib.h>
/* 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 beginging 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 beginging 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)
Output:
Given Linked List
1 2 3 4 5 6 7
Pre-Order Traversal of constructed BST
4 2 1 3 6 5 7
时间复杂度：O(n)
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