PAT 1099. Build A Binary Search Tree (30)

本文介绍了一种利用给定的节点信息构建二叉搜索树的方法,并通过层次遍历输出树的内容。首先根据节点信息建立二叉树结构,然后对节点值进行排序,再通过中序遍历将排序后的值填入树中。

这道题是二叉搜索树(BST)。
1.根据节点建树
2.把数组数排序
3.中序遍历二叉树,讲数据写入树中
4.层次遍历树

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;

struct p
{
  int left;
  int right;
  int value;
  bool flag;//是否已经有数字插入
}Node[100001];

int point[101];
int len = 0;

void insert(int i)
{
  if(Node[i].left != -1)
    insert(Node[i].left);
  Node[i].value = point[len++];
  if(Node[i].right != -1)
    insert(Node[i].right);
}

int main()
{
  int n;
  while(scanf("%d",&n) != EOF)
  {
     for(int i = 0; i < n; i++)
     {
       scanf("%d%d",&Node[i].left,&Node[i].right);
       Node[i].flag = false;//没有用到
     }
     for(int i = 0; i < n; i++)
       scanf("%d",&point[i]);
     sort(point,point+n);//排序
     insert(0);//遍历
     queue<int> q;
     q.push(0);
     while(!q.empty())
     {
       int x = q.front();
       q.pop();
       if(x != 0)
         printf(" ");
       printf("%d",Node[x].value);
       if(Node[x].left != -1)
         q.push(Node[x].left);
       if(Node[x].right != -1)
         q.push(Node[x].right);
     }
     printf("\n");
  }
  return 0;
}
【Solution】 To convert a binary search tree into a sorted circular doubly linked list, we can use the following steps: 1. Inorder traversal of the binary search tree to get the elements in sorted order. 2. Create a doubly linked list and add the elements from the inorder traversal to it. 3. Make the list circular by connecting the head and tail nodes. 4. Return the head node of the circular doubly linked list. Here's the Python code for the solution: ``` class Node: def __init__(self, val): self.val = val self.prev = None self.next = None def tree_to_doubly_list(root): if not root: return None stack = [] cur = root head = None prev = None while cur or stack: while cur: stack.append(cur) cur = cur.left cur = stack.pop() if not head: head = cur if prev: prev.right = cur cur.left = prev prev = cur cur = cur.right head.left = prev prev.right = head return head ``` To verify the accuracy of the code, we can use the following test cases: ``` # Test case 1 # Input: [4,2,5,1,3] # Output: # Binary search tree: # 4 # / \ # 2 5 # / \ # 1 3 # Doubly linked list: 1 <-> 2 <-> 3 <-> 4 <-> 5 # Doubly linked list in reverse order: 5 <-> 4 <-> 3 <-> 2 <-> 1 root = Node(4) root.left = Node(2) root.right = Node(5) root.left.left = Node(1) root.left.right = Node(3) head = tree_to_doubly_list(root) print("Binary search tree:") print_tree(root) print("Doubly linked list:") print_list(head) print("Doubly linked list in reverse order:") print_list_reverse(head) # Test case 2 # Input: [2,1,3] # Output: # Binary search tree: # 2 # / \ # 1 3 # Doubly linked list: 1 <-> 2 <-> 3 # Doubly linked list in reverse order: 3 <-> 2 <-> 1 root = Node(2) root.left = Node(1) root.right = Node(3) head = tree_to_doubly_list(root) print("Binary search tree:") print_tree(root) print("Doubly linked list:") print_list(head) print("Doubly linked list in reverse order:") print_list_reverse(head) ``` The output of the test cases should match the expected output as commented in the code.
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