• SDUST数据结构


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    函数题:

      6-1 求二叉树高度:

    裁判测试程序样例:

    #include <stdio.h>
    #include <stdlib.h>
    
    typedef char ElementType;
    typedef struct TNode *Position;
    typedef Position BinTree;
    struct TNode{
        ElementType Data;
        BinTree Left;
        BinTree Right;
    };
    
    BinTree CreatBinTree(); /* 实现细节忽略 */
    int GetHeight( BinTree BT );
    
    int main()
    {
        BinTree BT = CreatBinTree();
        printf("%d
    ", GetHeight(BT));
        return 0;
    }
    /* 你的代码将被嵌在这里 */

    代码:

    int GetHeight(BinTree BT)
    {
        if(BT == NULL)//判断是否为空 
        return 0;
        int l,r;
        l=GetHeight(BT->Left);//递归 
        r=GetHeight(BT->Right);
        return l>=r?l+1:r+1;    
    } 
    View Code

      6-2 二叉树的遍历:

    裁判测试程序样例:

    #include <stdio.h>
    #include <stdlib.h>
    
    typedef char ElementType;
    typedef struct TNode *Position;
    typedef Position BinTree;
    struct TNode{
        ElementType Data;
        BinTree Left;
        BinTree Right;
    };
    
    BinTree CreatBinTree(); /* 实现细节忽略 */
    void InorderTraversal( BinTree BT );
    void PreorderTraversal( BinTree BT );
    void PostorderTraversal( BinTree BT );
    void LevelorderTraversal( BinTree BT );
    
    int main()
    {
        BinTree BT = CreatBinTree();
        printf("Inorder:");    InorderTraversal(BT);    printf("
    ");
        printf("Preorder:");   PreorderTraversal(BT);   printf("
    ");
        printf("Postorder:");  PostorderTraversal(BT);  printf("
    ");
        printf("Levelorder:"); LevelorderTraversal(BT); printf("
    ");
        return 0;
    }
    /* 你的代码将被嵌在这里 */

    代码:

    void InorderTraversal( BinTree BT )//前三个依次递归,只需注意每一次的输出位置即可 
    {
        if(BT)
        {
            if(BT->Left)
                InorderTraversal(BT->Left);
            printf(" %c",BT->Data);
            if(BT->Right)
                InorderTraversal(BT->Right); 
        }
    }
    void PreorderTraversal( BinTree BT )
    {
        if(BT)
        {
            printf(" %c",BT->Data);
            if(BT->Left)
                PreorderTraversal(BT->Left);
            if(BT->Right)
                PreorderTraversal(BT->Right);
        }
    }
    void PostorderTraversal( BinTree BT )
    {
        if (BT)
        {
            if (BT->Left)
                 PostorderTraversal(BT->Left);
             if (BT->Right)
                 PostorderTraversal(BT->Right);
             printf(" %c", BT->Data);
        }
    }
    void LevelorderTraversal( BinTree BT )
    {
        BinTree bt[10001],root;
        int h=0,t=0;
        if(BT)
        {
            bt[t++]=BT;
            while(h!=t)
            {
                root=bt[h++];
                printf(" %c", root->Data);
                if(root->Left)
                    bt[t++] = root->Left;
                if(root->Right)
                    bt[t++] = root->Right;
            }
        } 
    }
    View Code

      6-3 先序输出叶结点:

    裁判测试程序样例:

    #include <stdio.h>
    #include <stdlib.h>
    
    typedef char ElementType;
    typedef struct TNode *Position;
    typedef Position BinTree;
    struct TNode{
        ElementType Data;
        BinTree Left;
        BinTree Right;
    };
    
    BinTree CreatBinTree(); /* 实现细节忽略 */
    void PreorderPrintLeaves( BinTree BT );
    
    int main()
    {
        BinTree BT = CreatBinTree();
        printf("Leaf nodes are:");
        PreorderPrintLeaves(BT);
        printf("
    ");
    
        return 0;
    }
    /* 你的代码将被嵌在这里 */

    代码:

    void PreorderPrintLeaves( BinTree BT )
    {
        if(BT==NULL)//若为空,只退回当前函数 
            return ;
        PreorderPrintLeaves(BT->Left);//递归调用 
        PreorderPrintLeaves(BT->Right);
        if(BT->Left==NULL&&BT->Right==NULL)
            printf(" %c",BT->Data);
        //PreorderPrintLeaves(BT->Left);//递归调用 
        //PreorderPrintLeaves(BT->Right);
    }
    View Code

      7-1 根据后序和中序遍历输出先序遍历:

    输入样例:

    7
    2 3 1 5 7 6 4
    1 2 3 4 5 6 7

    输出样例:

    Preorder: 4 1 3 2 6 5 7

    代码:

    #include<bits/stdc++.h>
    using namespace std;
    typedef struct BiTNode
    {
        int Data;
        struct BiTNode *lchild;
        struct BiTNode *rchild;
    }BiTNode, *BiTree;
    BiTree PlusTree(int *l, int *r, int n)
    {
        if(n==0)
            return NULL;
        else
        {
            BiTree t=new BiTNode;
            t->Data=r[n-1];
            int i=0;
            for(i=0;i<n;i++)
            {
                if(r[n-1]==l[i])
                    break;
            }
            t->lchild = PlusTree(l, r, i);
            t->rchild = PlusTree(l+i+1, r+i, n-i-1);
            return t;
        }
    }
    void PreorderTraversal(BiTree BT)
    {
        if(BT == NULL)
            return;
        printf(" %d", BT->Data);
        PreorderTraversal(BT->lchild);
        PreorderTraversal(BT->rchild);
    }
    int main()
    {
        int n;
        scanf("%d",&n);
        int a[35],b[35];
        for(int i=0;i<n;i++)
            cin>>b[i];
        for(int j=0;j<n;j++)
            cin>>a[j];
        BiTree tree;
        tree = PlusTree(a, b, n);
        printf("Preorder:");
        PreorderTraversal(tree);
        printf("
    ");
        return 0;
    }
     
    View Code

      7-2 树的同构:

    输入样例1:

    8
    A 1 2
    B 3 4
    C 5 -
    D - -
    E 6 -
    G 7 -
    F - -
    H - -
    8
    G - 4
    B 7 6
    F - -
    A 5 1
    H - -
    C 0 -
    D - -
    E 2 -

    输出样例:

    Yes

    输入样例:

    8
    B 5 7
    F - -
    A 0 3
    C 6 -
    H - -
    D - -
    G 4 -
    E 1 -
    8
    D 6 -
    B 5 -
    E - -
    H - -
    C 0 2
    G - 3
    F - -
    A 1 4

    输出样例:

    No

    代码:

    #include<cstdio>
    typedef int Tree;
    struct TNode{
        Tree left, right;
        char data;
        int flag;
    }T1[15], T2[15];
    int flag[15];
    Tree BuildTree( struct TNode T[]){
        Tree R=-1, cl, cr;
        int n;
        scanf("%d
    ", &n);
        for(int i=0; i<n; i++)flag[i]=0;
        for(int i=0; i<n; i++){
            scanf("%c %c %c
    ", &T[i].data, &cl, &cr);
            if(cl!='-'){
                T[i].left=cl - '0';
                flag[T[i].left]=1;
            }else{
                T[i].left=-1;
            }
            if(cr!='-'){
                T[i].right= cr-'0';
                flag[T[i].right]=1;
            }else{
                T[i].right=-1;
            }
        }
        for(int i=0; i<n; i++)
            if(flag[i]==0){
                R=i;
                break;
            }
        return R;
    }
    int PanDuan(Tree R1, Tree R2){
        if(R1==-1 && R2==-1)return 1;
        if((R1==-1 && R2!=-1) || (R2==-1 && R1!=-1))return 0;
        if(T1[R1].data!=T2[R2].data)return 0;
        if(T1[R1].left==-1 && T2[R2].left==-1)
            return PanDuan(T1[R1].right, T2[R2].right);
        if((T1[R1].left!=-1&&T2[R2].left!=-1)&&(T1[T1[R1].left].data==T2[T2[R2].left].data)){
            return(PanDuan(T1[R1].left, T2[R2].left)&&PanDuan(T1[R1].right, T2[R2].right));
        }else{
            return(PanDuan(T1[R1].right, T2[R2].left)&&PanDuan(T1[R1].left, T2[R2].right));
        }
    }
    int main(){
        Tree a, b;
        a=BuildTree(T1);
        b=BuildTree(T2);
        if(PanDuan(a, b)){
            printf("Yes");
        }else{
            printf("No");
        }
        return 0;
    }
    View Code

      7-3 修理牧场:

    输入样例:

    8
    4 5 1 2 1 3 1 1

    输出样例:

    49

    代码:

    #include<bits/stdc++.h>
    using namespace std;
    int main()
    {
        int n;
        scanf("%d",&n);
        priority_queue<int, vector<int>, greater<int> > w;
        int a[n];
        int i=0;
        int sum=0;   //待求最小值
        for(;i<n;i++)
        {
            scanf("%d",&a[i]);
            w.push(a[i]);
        }
        int x;//存取现小顶堆
        int y;//存取新小顶堆
        while(!w.empty())    //最短的木头二合一
        {
            x=w.top();
            w.pop();       //用完移除
    
            if(w.empty())  //移除x后没有元素了,说明x是木头总长,不再循环
            {
                break;
            }
            y=w.top();
            w.pop();
            x+=y;
            sum+=x;
            w.push(x);   //新木头放入堆中
        }
        printf("%d",sum);
    }
    View Code

      7-4 完全二叉搜索树:

    输入样例:

    10
    1 2 3 4 5 6 7 8 9 0

    输出样例:

    6 3 8 1 5 7 9 0 2 4

    代码:

    #include<bits/stdc++.h>
    using namespace std;
    int n, flag;
    int a[1001];
    int b[1001];
    void DFS(int aa)
    {
        if(aa>n)
            return ;
        DFS(2*aa);
        b[aa]=++flag;
        DFS(2*aa+1);
    }
    int main()
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        sort(a+1, a+n+1);
        DFS(1);
        for(int i=1;i<=n;i++)
        {
            if(i==1)
                printf("%d",a[b[i]]);
            else
                printf(" %d",a[b[i]]);
        }
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/3cH0-Nu1L/p/14045086.html
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