1099. Build A Binary Search Tree (30)

本文介绍了一个PAT-A 1099题目的解决方案,该题目要求构建一棵二叉树并进行层次遍历。文章详细展示了如何通过输入的节点关系和权值构建有序的二叉树,并通过层次遍历输出结果。

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题目详情: https://www.patest.cn/contests/pat-a-practise/1099

提交:这里写图片描述

提交代码:

#include <iostream>
#include <algorithm>
using namespace std;
#define N 110
typedef struct Node
{
    int data;
    int leftChild;
    int rightChild;
}Node;
Node node[N]; //nodes[] are numbered from 0 to n-1,and 0 is always the root
int n,weight[N];
int queue[N],front,rear,level[N],top = -1;
int getChildNumber( int root,int &number)  //得到以root为根节点的树中节点的个数,包括该根节点 
{
    if( root != -1 )
    {
        ++number;
        getChildNumber(node[root].leftChild,number);
        getChildNumber(node[root].rightChild,number);
        return number;
    }
    return 0;
}
void build( int root,int *weight )
{
    if( root == -1 ) return ;  //没有节点直接返回 
    int leftNumber = 0,rightNumber = 0;
    leftNumber = getChildNumber(node[root].leftChild,leftNumber);  //得到左子树的节点个数 
    rightNumber = getChildNumber(node[root].rightChild,rightNumber); //得到右子树的节点个数 
    cout<<"leftNumber is "<<leftNumber<<",rightNumber is "<<rightNumber<<endl;
    cout<<"will insert "<<weight[leftNumber]<<endl;9
    node[root].data = weight[leftNumber];
    build(node[root].leftChild,weight); //递归左子树 ,同时改变数组的起始地址,若不改会出现问题 
    build(node[root].rightChild,weight+leftNumber+1); //递归右子树,同时改变数组的起始地址 
}
void levelOrder( int start )
{
    queue[++rear] = start;
    while( front != rear )
    {
        int key = queue[++front];
        if( node[key].leftChild != -1 )
        {
            queue[++rear] = node[key].leftChild;    
        }
        if( node[key].rightChild != -1 )
        {
            queue[++rear] = node[key].rightChild;
        }
        level[++top] = node[key].data;
    }
}
int main ()
{
    cin>>n;
    for( int i=0;i<n;++i )
    {
        int leftKid,rightKid;
        cin>>leftKid>>rightKid;
        node[i].leftChild = leftKid;
        node[i].rightChild = rightKid;
    }
    for( int i=0;i<n;++i )
        cin>>weight[i];
    sort(weight,weight+n); //使weight数组有序 ,以便建立二叉树时使用 
    build(0,weight);       //建立该二叉树 
    rear = front = -1; //层次遍历前的初始化操作 
    levelOrder(0); //进行层次遍历 
    for( int i=0;i<=top;++i ) //输出层次遍历的序列 
    {
        if( i == rear )
            cout<<level[i]<<endl;
        else
            cout<<level[i]<<" ";
    }
    return 0;
}

虽然通过了测试但是有个地方搞不懂,就是在getChildNumber()函数中,如果不加最后一句代码,即:return 0;会返回-1,这个地方不明白,希望得到大神指点!

【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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