c语言将链表写入二进制文件

Problem statement: Write a C program to convert a binary tree into a single linked list by traversing level-wise.

问题陈述：编写一个C程序，通过逐级遍历将二进制树转换为单个链表 。

Example:

例：

The above binary tree is converted to 2 → 7 → 5 → 2 → 6 → 9 → 5 → 11 → 4 → NULL

上面的二叉树被转换为2→7→5→2→6→9→5→11→4→NULL

Solution

解

1. Building a linked list - Setting the first node as Head and then appending other nodes.

   构建链接列表 -将第一个节点设置为Head，然后附加其他节点。

2. Traversing & displaying a single link list.

   遍历并显示单个链接列表 。

3. Building a tree & level order traversal of a tree.

   建立树和树的层级遍历 。

Algorithm:

算法：

1. Assign the root value as head node value in linked list.

   在链接列表中将根值分配为头节点值。

2. Do level-order traversal

   进行水平顺序遍历

   For each

   对于每个

   tree node create a single linked list node and append it to the linked list.

   树节点创建一个链接列表节点 ，并将其附加到链接列表。

3. Display the single linked list.

   显示单个链接列表。

How the tree is converted to the single list...

树如何转换为单个列表...

Let’s do solve the above example.

让我们来解决上面的例子。

Root=2;
Queue status: 2
----------------------------------------------------
1st iteration
Queue not empty
Queue front is 2
Head is null, thus head is 2
Linked list up to now:2->NULL
Pop 2
Push: 2->left(7) & 2->right(5)
Queue status: 7, 5
----------------------------------------------------
2nd iteration
Queue not empty
Queue front is 7
Head is not null, thus append 7
Linked list up to now:2->7->NULL
Pop 7
Push: 7->left(2)& 7->right(6)
Queue status: 5, 2, 6
----------------------------------------------------
3rd iteration
Queue not empty
Queue front is 5
Head is not null, thus append 5
Linked list up to now:2->7->5->NULL
Pop 5
Push: 5->right (9) only (5->left is NULL)
Queue status: 2, 6, 9 
----------------------------------------------------
4th iteration
Queue not empty
Queue front is 2
Head is not null, thus append 2
Linked list up to now:2->7->5->2->NULL
Pop 2
Push: Nothing ( both child are NULL)
Queue status: 6, 9 
----------------------------------------------------
5th iteration
Queue not empty
Queue front is 6
Head is not null, thus append 6
Linked list up to now:2->7->5->2->6->NULL
Pop 6
Push: 6->left(5) and 6->right(11)
Queue status: 9, 5, 11
----------------------------------------------------
6th iteration
Queue not empty
Queue front is 9
Head is not null, thus append 9
Linked list up to now:2->7->5->2->6->9->NULL
Pop 9
Push: 9->left(4) only (right child NULL)
Queue status: 5, 11, 4
----------------------------------------------------
7th iteration
Queue not empty
Queue front is 5
Head is not null, thus append 5
Linked list up to now:2->7->5->2->6->9->5->NULL
Pop 5
Push: Nothing (both child NULL)
Queue status: 11, 4 
----------------------------------------------------
8th iteration
Queue not empty
Queue front is 11
Head is not null, thus append 11
Linked list up to now:2->7->5->2->6->9->5->11->NULL
Pop 11
Push: Nothing (both child NULL)
Queue status: 4 
----------------------------------------------------
8th iteration
Queue not empty
Queue front is 4
Head is not null, thus append 4
Linked list up to now: 2->7->5->2->6->9->5->11->4->NULL
Pop 4
Push: Nothing (both child NULL)
Queue status: empty queue
----------------------------------------------------
Iteration stops
So final link list is: 2->7->5->2->6->9->5->11->4->NULL

通过逐级遍历将二叉树转换为单链接列表的C实现 (C implementation to to convert a Binary Tree into a Singly Linked List by Traversing Level by Level)

#include <bits/stdc++.h>
using namespace std;
// tree node is defined
class tree{    
public:
int data;
tree *left;
tree *right;
};
class sll{
public:
int data;
sll* next;
};
sll* creatnode(int d){ //create node for single linked list
sll* temp=(sll*)malloc(sizeof(sll));
temp->data=d;
temp->next=NULL;
return temp;
}
void display(sll* head){
sll* current=head; // current node set to head
printf("displayig the converted list...\n");
while(current!=NULL){ //traverse until current node isn't NULL
if(current->next)
printf("%d->",current->data);
else
printf("%d->NULL\n",current->data);
current=current->next; // go to next node
}
}
sll* flatten(tree* root)
{
//Declare queue using STL 
sll* head=NULL,*tempL;
queue<tree*> q;
//enqueue the root
q.push(root);
vector<int> store;
tree* temp;
//do the level order traversal & build single linked list
while(!q.empty()){
//dequeue
temp=q.front();
q.pop();
if(head==NULL){//for inserting first node
head=creatnode(temp->data);
tempL=head;
}
else{//for inserting rest of the nodes
tempL->next=creatnode(temp->data);
tempL=tempL->next;
}
// do level order traversing
if(temp->left)//for left child
q.push(temp->left);
if(temp->right)//for right child
q.push(temp->right);
}
return head;
}
tree* newnode(int data)  // creating new node for tree
{ 
tree* node = (tree*)malloc(sizeof(tree)); 
node->data = data; 
node->left = NULL; 
node->right = NULL; 
return(node); 
} 
int main() 
{ 
//**same tree is builted as shown in example**
cout<<"same tree is built as shown in example\n";
tree *root=newnode(2); 
root->left= newnode(7); 
root->right= newnode(5); 
root->right->right=newnode(9);
root->right->right->left=newnode(4);
root->left->left=newnode(2); 
root->left->right=newnode(6);
root->left->right->left=newnode(5);
root->left->right->right=newnode(11);
cout<<"converting the tree into a single link list...\n";
cout<<"by traversing the tree level-wise\n";
sll* head=flatten(root);
//displaying the list built from the tree
display(head);
return 0; 
} 

Output

输出量

same tree is built as shown in example 
converting the tree into a single link list...
by traversing the tree level-wise
displayig the converted list...
2->7->5->2->6->9->5->11->4->NULL

翻译自: https://www.includehelp.com/c-programs/convert-a-binary-tree-into-a-singly-linked-list-by-traversing-level-by-level.aspx

c语言将链表写入二进制文件