リーフノードのリスト(PAT 3-2)
4196 ワード
参照先:http://www.patest.cn/contests/mooc-ds/03-2
質問:
時間の制限
400 ms
メモリ制限
65536 kB
コード長制限
8000 B
クイズルーチン
Standard
作成者
CHEN, Yue
Given a tree, you are supposed to list all the leaves in the order of top down, and left to right.
Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N (<=10) which is the total number of nodes in the tree -- and hence the nodes are numbered from 0 to N-1. Then N lines follow, each corresponds to a node, and gives the indices of the left and right children of the node. If the child does not exist, a "-"will be put at the position. Any pair of children are separated by a space.
Output Specification:
For each test case, print in one line all the leaves' indices in the order of top down, and left to right. There must be exactly one space between any adjacent numbers, and no extra space at the end of the line.
Sample Input:(入力に順序がなく、各行がサブノードインデックス、例えば2 7はこのノードの左のサブノードが2行目のノード(0-)、右のサブノードが7行目のノード(4 6)であることを示す)
Sample Output:
実際のツリー構造は次のとおりです.
#include "stdafx.h"#include #include #define MAXSIZE 16 typedef struct TNode{ struct TNode *left; struct TNode * right; int isFirst; //isRoot node int data; }TNode; typedef struct Queue { TNode* node[MAXSIZE]; int top ,bottom; int size; }Queue; Queue* initQueue(Queue* queue) {queue->top=queue->bottom=0; //最初は0位置queue->size=MAXSIZE;//printf("bottom=%d",Queue->top);return queue;}void push(Queue*queue,Tnode*p){ queue->node[queue->bottom]=p; queue->bottom++; } TNode* pop(Queue* queue) //p is a pointer to pointer ,nice design { TNode* ret=queue->node[queue->top]; queue->top++; return ret; } bool isEmpty(Queue* queue){ if(queue->top==queue->bottom) return true; return false; } TNode * list[MAXSIZE]; TNode * inputTNode(TNode *bt) {if(bt==NULL) bt=(TNode*)malloc(sizeof(TNode)); char s[3],u[3]; scanf("%s%s",u,s); if(s[0]=='-') bt->right=NULL; else{bt->right=(TNode*)malloc(sizeof(TNode)); bt->right->data=atoi(s);} if(u[0]=='-') bt->left=NULL; else{bt->left=(TNode*)malloc(sizeof(TNode)); bt->left->data=atoi(u);} return bt; } TNode * initBTree() { TNode* bt=NULL; int num,i; scanf("%d",&num); if(num==-1) return NULL; for(i=0;i {bt=(TNode*)malloc(sizeof(TNode)); list[i]=bt; bt=inputTNode(bt); } for(i=0;i {TNode* bt=NULL;bt=list[i];//printf("bt value %d",bt->data);if(bt->left){int tmp=bt->left->data;bt->left=list[tmp];bt->left->data=tmp;printf("bt value %d",bt->left->data);bt->left->isFirst=1;}if(bt->right){int tmp=bt->right->data; bt->right=list[bt->right->data];bt->right->data=tmp;bt->right->isFirst=1;printf("bt value %d",bt->right->data);} } TNode* root=NULL; for(i=0;i {root=list[i];if(root->isFirst!=1) break; } return root; } int main(int argc, char* argv[]) {//Queue.top=-1; Queue queue;Queue *q=&queue;q=initQueue(q);TNode * root=initBTree(); push(q,root); int first=1; while(!isEmpty(q)) { TNode * tmp=pop(q); if((tmp->right==NULL)&&(tmp->left==NULL)) { if(first) { first=0; printf("%d",tmp->data); } else printf("%d",tmp->data); } if(tmp->left) push(q,tmp->left); if(tmp->right) push(q,tmp->right); }return 0; }
質問:
時間の制限
400 ms
メモリ制限
65536 kB
コード長制限
8000 B
クイズルーチン
Standard
作成者
CHEN, Yue
Given a tree, you are supposed to list all the leaves in the order of top down, and left to right.
Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N (<=10) which is the total number of nodes in the tree -- and hence the nodes are numbered from 0 to N-1. Then N lines follow, each corresponds to a node, and gives the indices of the left and right children of the node. If the child does not exist, a "-"will be put at the position. Any pair of children are separated by a space.
Output Specification:
For each test case, print in one line all the leaves' indices in the order of top down, and left to right. There must be exactly one space between any adjacent numbers, and no extra space at the end of the line.
Sample Input:(入力に順序がなく、各行がサブノードインデックス、例えば2 7はこのノードの左のサブノードが2行目のノード(0-)、右のサブノードが7行目のノード(4 6)であることを示す)
8
1 -
- -
0 -
2 7
- -
- -
5 -
4 6
Sample Output:
4 1 5実際のツリー構造は次のとおりです.
#include "stdafx.h"#include #include #define MAXSIZE 16 typedef struct TNode{ struct TNode *left; struct TNode * right; int isFirst; //isRoot node int data; }TNode; typedef struct Queue { TNode* node[MAXSIZE]; int top ,bottom; int size; }Queue; Queue* initQueue(Queue* queue) {queue->top=queue->bottom=0; //最初は0位置queue->size=MAXSIZE;//printf("bottom=%d",Queue->top);return queue;}void push(Queue*queue,Tnode*p){ queue->node[queue->bottom]=p; queue->bottom++; } TNode* pop(Queue* queue) //p is a pointer to pointer ,nice design { TNode* ret=queue->node[queue->top]; queue->top++; return ret; } bool isEmpty(Queue* queue){ if(queue->top==queue->bottom) return true; return false; } TNode * list[MAXSIZE]; TNode * inputTNode(TNode *bt) {if(bt==NULL) bt=(TNode*)malloc(sizeof(TNode)); char s[3],u[3]; scanf("%s%s",u,s); if(s[0]=='-') bt->right=NULL; else{bt->right=(TNode*)malloc(sizeof(TNode)); bt->right->data=atoi(s);} if(u[0]=='-') bt->left=NULL; else{bt->left=(TNode*)malloc(sizeof(TNode)); bt->left->data=atoi(u);} return bt; } TNode * initBTree() { TNode* bt=NULL; int num,i; scanf("%d",&num); if(num==-1) return NULL; for(i=0;i {bt=(TNode*)malloc(sizeof(TNode)); list[i]=bt; bt=inputTNode(bt); } for(i=0;i {TNode* bt=NULL;bt=list[i];//printf("bt value %d",bt->data);if(bt->left){int tmp=bt->left->data;bt->left=list[tmp];bt->left->data=tmp;printf("bt value %d",bt->left->data);bt->left->isFirst=1;}if(bt->right){int tmp=bt->right->data; bt->right=list[bt->right->data];bt->right->data=tmp;bt->right->isFirst=1;printf("bt value %d",bt->right->data);} } TNode* root=NULL; for(i=0;i {root=list[i];if(root->isFirst!=1) break; } return root; } int main(int argc, char* argv[]) {//Queue.top=-1; Queue queue;Queue *q=&queue;q=initQueue(q);TNode * root=initBTree(); push(q,root); int first=1; while(!isEmpty(q)) { TNode * tmp=pop(q); if((tmp->right==NULL)&&(tmp->left==NULL)) { if(first) { first=0; printf("%d",tmp->data); } else printf("%d",tmp->data); } if(tmp->left) push(q,tmp->left); if(tmp->right) push(q,tmp->right); }return 0; }