PAT 1043. Is It a Binary Search Tree

题目:http://pat.zju.edu.cn/contests/pat-a-practise/1043

题解:

1:通过输入构造二叉排序树

2:先序遍历并与输入顺序比较

3:如果比较不符合再构造镜像二叉排序树

4:比较输出结果

代码:

#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<algorithm>
using namespace std;
#define INF 0x6fffffff
#define MAX 1005
struct point
{
    int value;
    int left;
    int right;
} node[MAX];
int a[MAX],b[MAX];
int deep;
void dfs(int x,int idx)
{//构造二叉树
    if(node[x].value>node[idx].value)
    {
        //应在当前结点左子树
        if(node[x].left==-1)
            node[x].left=idx;
        else
            dfs(node[x].left,idx);
    }
    else
    {
        //应在当前结点右子树
        if(node[x].right==-1)
            node[x].right=idx;
        else
            dfs(node[x].right,idx);
    }
}
void dfsP(int x,int idx)
{//构造镜像排序二叉树
    if(node[x].value>node[idx].value)
    {
        //应在当前结点右子树
        if(node[x].right==-1)
            node[x].right=idx;
        else
            dfsP(node[x].right,idx);
    }
    else
    {
        //应在当前结点左子树
        if(node[x].left==-1)
            node[x].left=idx;
        else
            dfsP(node[x].left,idx);
    }
}
void dfsPre(int x)
{//先序遍历
    b[deep]=node[x].value;
    ++deep;
    if(node[x].left!=-1)
        dfsPre(node[x].left);
    if(node[x].right!=-1)
        dfsPre(node[x].right);
}
void dfsPost(int x)
{//后序遍历
    if(node[x].left!=-1)
        dfsPost(node[x].left);
    if(node[x].right!=-1)
        dfsPost(node[x].right);
    b[deep]=node[x].value;
    ++deep;
}
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=0; i<n; ++i)
    {
        scanf("%d",&node[i].value);
        a[i]=node[i].value;
        node[i].left=node[i].right=-1;
    }
    for(int i=1; i<n; ++i)
    {//构造排序二叉树
        dfs(0,i);
    }
    deep=0;
    dfsPre(0);//先序遍历
    bool flag=true;
    for(int i=0; i<n; ++i)
    {//比较
        if(a[i]!=b[i])
        {
            flag=false;
            break;
        }
    }
    if(flag)//先序遍历与输入相同
    {
        printf("YES\n");
        deep=0;
        dfsPost(0);//后序遍历
        bool first=true;
        for(int i=0; i<n; ++i)
        {
            if(first)
            {
                printf("%d",b[i]);
                first=false;
            }
            else
                printf(" %d",b[i]);
        }
        printf("\n");
    }
    else
    {
        for(int i=0; i<n; ++i)
            node[i].left=node[i].right=-1;
        for(int i=1; i<n; ++i)
        {//构造镜像二叉排序树
            dfsP(0,i);
        }
        deep=0;
        dfsPre(0);
        flag=true;
        for(int i=0; i<n; ++i)
        {//比较
            if(a[i]!=b[i])
            {
                flag=false;
                break;
            }
        }
        if(flag)//镜像二叉排序树的先序遍历与输入相同
        {
            printf("YES\n");
            deep=0;
            dfsPost(0);
            bool first=true;
            for(int i=0; i<n; ++i)
            {
                if(first)
                {
                    printf("%d",b[i]);
                    first=false;
                }
                else
                    printf(" %d",b[i]);
            }
            printf("\n");
        }
        else
            printf("NO\n");
    }
    return 0;
}

来源: http://blog.youkuaiyun.com/acm_ted/article/details/20715415


【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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