• PAT-1099(Build A Binary Search Tree)


      题目见这里

          (分析) 分四步进行:

          1)根据给定的结点情况建二叉树  2)对输入的键值排序(asending) 3)对二叉树中序遍历,同时对应赋key值 4)层次遍历(队列应用)

          题目并不困难,但是我误入了trick,错误假定了结点按先序遍历是按顺序编号的(当然是受样例的影响),所以有了下面22分(满分30) 的submit(贴出来是因为这不失为好的巩固二叉树知识的程序)

    #include <stdio.h> 
    #include <stdlib.h> //qsort,malloc
    #define N 105
    
    typedef struct node{
    	int data;
    	struct node *lChild,*rChild;
    }BiNode,*BiTree;
    
    void CreatBiTree(BiTree *bt){
    	(*bt) = (BiTree)malloc(sizeof(BiNode));
    	(*bt)->lChild = (*bt)->rChild = NULL;
    	int left,right;
    	scanf("%d%d",&left,&right);
    	if(left!=-1) CreatBiTree(&((*bt)->lChild));
    	if(right!=-1) CreatBiTree(&((*bt)->rChild)); 
    }
    
    
    int cmp(const void *a, const void *b){ //asending 
    	return *(int *)a - *(int *)b;
    }
    
    void Sort(int *key, int n){
    	int i;
    	for(i=0;i<n;i++)
    		scanf("%d",&key[i]);
    	qsort((void*)key,n,sizeof(key[0]),cmp);
    }
    
    void LDR(BiTree bt, int key[]){
    	static int i = 0;
    	if(bt){
    		LDR(bt->lChild,key);
    		bt->data = key[i++];
    		LDR(bt->rChild,key);
    	}
    }
    
    void LOT(BiTree bt){
    	BiTree q[N],bNode;
    	int front,rear,flag;
    	rear = front = flag = 0;
    	q[rear] = bt;
    	while(front<=rear){
    		bNode = q[front++];
    		if(!flag){
    			printf("%d",bNode->data);
    			flag = 1;
    		}
    		else printf(" %d",bNode->data);
    		if(bNode->lChild) q[++rear] = bNode->lChild;
    		if(bNode->rChild) q[++rear] = bNode->rChild;
    	}
    	printf("
    ");
    }
    
    void DestryBiTree(BiTree *bt){
    	if(*bt){
    		DestryBiTree(&((*bt)->lChild));
    		DestryBiTree(&((*bt)->rChild));
    		free(*bt);
    	}
    }
    
    int main(){
    	int key[N];
    	BiTree bt;
    	int n;	
    //	freopen("Data.txt","r",stdin);
    	scanf("%d",&n);
    	CreatBiTree(&bt,n);
    	Sort(key,n);
    	LDR(bt,key); //中序遍历 
    	LOT(bt); //层次遍历 
    	DestryBiTree(&bt);	
    	return 0;
    }
    

      当然,既然认识到这个错误,当然是因为找到了反例:

           8
      7 1
      2 3
      -1 -1
      -1 4
      5 6
      -1 -1
      -1 -1
      -1 -1
      33 37 34 30 50 43 37 33

          从上面的反例中,我们注意到创建链表形式的二叉树是不太可能的,而应采用数组形式,所以有了AC的提交:

    #include <stdio.h> 
    #define N 105
    
    typedef struct{
    	int lChild,rChild;
    	int data;
    }Node;
    
    void CreatBiTree(Node node[], int n){
    	int i = 0;
    	for(;i<n;i++) scanf("%d%d",&node[i].lChild,&node[i].rChild);
    }
    
    int cmp(const void *a, const void *b){ //asending 
    	return *(int *)a - *(int *)b;
    }
    
    void Sort(int *key, int n){
    	int i;
    	for(i=0;i<n;i++)
    		scanf("%d",&key[i]);
    	qsort((void*)key,n,sizeof(key[0]),cmp);
    }
    
    void LDR(Node *node, int key[], int i){
    	static int j = 0;
    	if(node[i].lChild!=-1) LDR(node,key,node[i].lChild);
    	node[i].data = key[j++];
    	if(node[i].rChild!=-1) LDR(node,key,node[i].rChild);//注意不要写误 
    }
    
    void LOT(Node node[]){
    	Node q[N],qNode;
    	int front,rear;
    	rear = front = 0;
    	q[rear] = node[0];
    	while(front<=rear){
    		qNode = q[front++];
    		if(rear) printf(" ");
    		printf("%d",qNode.data);
    		if(qNode.lChild!=-1) q[++rear] = node[qNode.lChild];
    		if(qNode.rChild!=-1) q[++rear] = node[qNode.rChild];
    	}
    	printf("
    ");
    }
    
    int main(){
    	int key[N];
    	Node node[N];
    	int n;	
    //	freopen("Data.txt","r",stdin);
    	scanf("%d",&n);
    	CreatBiTree(node,n);
    	Sort(key,n);
    	LDR(node,key,0); //中序遍历
    	LOT(node); //层次遍历 
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/emptyCoder/p/7134542.html
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