从给定的二叉树创建一个排序的链表
给定一棵二叉树，任务是将其转换为有序链表。
例子：
Input:
1
/ \
2 3
Output: 1 2 3
Input:
2
/ \
4 8
/ \ / \
7 3 5 1
Output: 1 2 3 4 5 7 8
Input:
3
/
4
/
1
/
9
Output: 1 3 4 9方法：递归迭代给定的二叉树，并使用插入排序将每个节点添加到结果链表（最初为空）中的正确位置。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// A linked list node
class Node {
public:
int data;
Node* next;
Node(int data)
{
this->data = data;
this->next = NULL;
}
};
// A binary tree node
class treeNode {
public:
int data;
treeNode* left;
treeNode* right;
treeNode(int data)
{
this->data = data;
this->left = NULL;
this->right = NULL;
}
};
// Function to print the linked list
void print(Node* head)
{
if (head == NULL) {
return;
}
Node* temp = head;
while (temp != NULL) {
cout << temp->data << " ";
temp = temp->next;
}
}
// Function to create Linked list from given binary tree
Node* sortedList(Node* head, treeNode* root)
{
// return head if root is null
if (root == NULL) {
return head;
}
// First make the sorted linked list
// of the left sub-tree
head = sortedList(head, root->left);
Node* newNode = new Node(root->data);
Node* temp = head;
Node* prev = NULL;
// If linked list is empty add the
// node to the head
if (temp == NULL) {
head = newNode;
}
else {
// Find the correct position of the node
// in the given linked list
while (temp != NULL) {
if (temp->data > root->data) {
break;
}
else {
prev = temp;
temp = temp->next;
}
}
// Given node is to be attached
// at the end of the list
if (temp == NULL) {
prev->next = newNode;
}
else {
// Given node is to be attached
// at the head of the list
if (prev == NULL) {
newNode->next = temp;
head = newNode;
}
else {
// Insertion in between the list
newNode->next = temp;
prev->next = newNode;
}
}
}
// Now add the nodes of the right sub-tree
// to the sorted linked list
head = sortedList(head, root->right);
return head;
}
// Driver code
int main()
{
/* Tree:
10
/ \
15 2
/ \
1 5
*/
treeNode* root = new treeNode(10);
root->left = new treeNode(15);
root->right = new treeNode(2);
root->left->left = new treeNode(1);
root->left->right = new treeNode(5);
Node* head = sortedList(NULL, root);
print(head);
return 0;
} Java
// Java implementation of the approach
import java.util.*;
class GFG
{
// A linked list node
static class Node
{
int data;
Node next;
Node(int data)
{
this.data = data;
this.next = null;
}
};
// A binary tree node
static class treeNode
{
int data;
treeNode left;
treeNode right;
treeNode(int data)
{
this.data = data;
this.left = null;
this.right = null;
}
};
// Function to print the linked list
static void print(Node head)
{
if (head == null)
{
return;
}
Node temp = head;
while (temp != null)
{
System.out.print(temp.data + " ");
temp = temp.next;
}
}
// Function to create Linked list from given binary tree
static Node sortedList(Node head, treeNode root)
{
// return head if root is null
if (root == null)
{
return head;
}
// First make the sorted linked list
// of the left sub-tree
head = sortedList(head, root.left);
Node newNode = new Node(root.data);
Node temp = head;
Node prev = null;
// If linked list is empty add the
// node to the head
if (temp == null)
{
head = newNode;
}
else
{
// Find the correct position of the node
// in the given linked list
while (temp != null)
{
if (temp.data > root.data)
{
break;
}
else
{
prev = temp;
temp = temp.next;
}
}
// Given node is to be attached
// at the end of the list
if (temp == null)
{
prev.next = newNode;
}
else
{
// Given node is to be attached
// at the head of the list
if (prev == null)
{
newNode.next = temp;
head = newNode;
}
else
{
// Insertion in between the list
newNode.next = temp;
prev.next = newNode;
}
}
}
// Now add the nodes of the right sub-tree
// to the sorted linked list
head = sortedList(head, root.right);
return head;
}
// Driver code
public static void main(String[] args)
{
/* Tree:
10
/ \
15 2
/ \
1 5
*/
treeNode root = new treeNode(10);
root.left = new treeNode(15);
root.right = new treeNode(2);
root.left.left = new treeNode(1);
root.left.right = new treeNode(5);
Node head = sortedList(null, root);
print(head);
}
}
// This code is contributed by 29AjayKumarC#
// C# implementation of the approach
using System;
class GFG
{
// A linked list node
class Node
{
public int data;
public Node next;
public Node(int data)
{
this.data = data;
this.next = null;
}
};
// A binary tree node
class treeNode
{
public int data;
public treeNode left;
public treeNode right;
public treeNode(int data)
{
this.data = data;
this.left = null;
this.right = null;
}
};
// Function to print the linked list
static void print(Node head)
{
if (head == null)
{
return;
}
Node temp = head;
while (temp != null)
{
Console.Write(temp.data + " ");
temp = temp.next;
}
}
// Function to create Linked list
// from given binary tree
static Node sortedList(Node head, treeNode root)
{
// return head if root is null
if (root == null)
{
return head;
}
// First make the sorted linked list
// of the left sub-tree
head = sortedList(head, root.left);
Node newNode = new Node(root.data);
Node temp = head;
Node prev = null;
// If linked list is empty add the
// node to the head
if (temp == null)
{
head = newNode;
}
else
{
// Find the correct position of the node
// in the given linked list
while (temp != null)
{
if (temp.data > root.data)
{
break;
}
else
{
prev = temp;
temp = temp.next;
}
}
// Given node is to be attached
// at the end of the list
if (temp == null)
{
prev.next = newNode;
}
else
{
// Given node is to be attached
// at the head of the list
if (prev == null)
{
newNode.next = temp;
head = newNode;
}
else
{
// Insertion in between the list
newNode.next = temp;
prev.next = newNode;
}
}
}
// Now add the nodes of the right sub-tree
// to the sorted linked list
head = sortedList(head, root.right);
return head;
}
// Driver code
public static void Main(String[] args)
{
/* Tree:
10
/ \
15 2
/ \
1 5
*/
treeNode root = new treeNode(10);
root.left = new treeNode(15);
root.right = new treeNode(2);
root.left.left = new treeNode(1);
root.left.right = new treeNode(5);
Node head = sortedList(null, root);
print(head);
}
}
// This code is contributed by PrinciRaj1992Python3
# Python3 implementation of the approach
# A linked list node
class Node:
def __init__(self, data = 0):
self.data = data
self.next = None
# A binary tree node
class treeNode:
def __init__(self, data):
self.data = data
self.left = None
self.right = None
# Function to print the linked list
def print_(head):
if (head == None):
return
temp = head
while (temp != None):
print ( temp.data, end = " " )
temp = temp.next
# Function to create Linked list from given binary tree
def sortedList( head, root):
# return head if root is None
if (root == None) :
return head
# First make the sorted linked list
# of the left sub-tree
head = sortedList(head, root.left)
newNode = Node(root.data)
temp = head
prev = None
# If linked list is empty add the
# node to the head
if (temp == None) :
head = newNode
else:
# Find the correct position of the node
# in the given linked list
while (temp != None):
if (temp.data > root.data) :
break
else:
prev = temp
temp = temp.next
# Given node is to be attached
# at the end of the list
if (temp == None):
prev.next = newNode
else:
# Given node is to be attached
# at the head of the list
if (prev == None) :
newNode.next = temp
head = newNode
else:
# Insertion in between the list
newNode.next = temp
prev.next = newNode
# Now add the nodes of the right sub-tree
# to the sorted linked list
head = sortedList(head, root.right)
return head
# Driver code
# Tree:
# 10
# / \
# 15 2
# / \
#1 5
root = treeNode(10)
root.left = treeNode(15)
root.right = treeNode(2)
root.left.left = treeNode(1)
root.left.right = treeNode(5)
head = sortedList(None, root)
print_(head)
# This code is contributed by Arnab KunduJavascript
输出：
1 2 5 10 15时间复杂度： O(n 2 )