[pta] 03 - 나무 2 List Leaves (25 점)
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 NN (\le 10≤10) which is the total number of nodes in the tree – and hence the nodes are numbered from 0 to N-1N−1. Then NN 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:
8 1 - - - 0 - 2 7 - - - - 5 - 4 6
Sample Output:
4 1 5
층 차 를 옮 겨 다 니 는 방법 을 사용 하 다.
#include
#include
#define MaxTree 10
#define ElementType int
#define Tree int
#define Null -1
struct TreeNode{
ElementType Element;
Tree Left;
Tree Right;
}T[MaxTree];
int BuildTree(){
int i,N;
char cl,cr;
//printf("enter N:");
scanf("%d",&N);
InitializeTree(N);
ReadTree(N);
return N;
//initial Tree
}
void InitializeTree(int N){
struct TreeNode *TreePointer;
for(int i=0;iElement=i;
TreePointer->Left=-1;
TreePointer->Right=-1;
//printf("e:%d l:%d r:%d
",TreePointer->Element,TreePointer->Left,TreePointer->Right);
}
}
//read Tree
void ReadTree(int N){
char cl,cr;
int i;
struct TreeNode *TreePointer;
for(i=0;i//printf("enter[%d] left & right :",i);
scanf(" %c",&cl);
scanf(" %c",&cr);
if(cl!='-'){
TreePointer->Left=cl-'0';
}
if(cr!='-'){
TreePointer->Right=cr-'0';
}
}
return 0;
}
void PrintTree(int N){
for(int i=0;iprintf("e:%d l:%d r:%d
",T[i].Element,T[i].Left,T[i].Right);
}
int findRoot(int N){
int i,j,root;
root = 0;
for(i=0;ifor(j=0;j//printf("T[%d].Left:%d T[%d].Right:%d T[%d].Element:%d
",j,T[j].Left,j,T[j].Right,root,T[root].Element);
if(T[j].Left==T[root].Element||T[j].Right==T[root].Element){
//printf("root changed");
root = j;
break;
}
}
}
return root;
}
void Leaves(int N,int root){
int a[N],i,j=0;
a[0]=root;
int flag=0;
for(int i=0;i//printf("a[%d] :%d
",i,a[i]);
//printf("e:%d l:%d r:%d
",T[a[i]].Element,T[a[i]].Left,T[a[i]].Right);
if(T[a[i]].Left!=-1){
j++;
a[j]=T[a[i]].Left;
}
if(T[a[i]].Right!=-1){
j++;
a[j]=T[a[i]].Right;
}
if(T[a[i]].Left==-1&&T[a[i]].Right==-1){
if(flag==0){
printf("%d",a[i]);
flag=1;
}else printf(" %d",a[i]);
}
}
}
int main(){
int root,N;
N = BuildTree(&T);
//PrintTree(N);
root=findRoot(N);
//printf("root:%d
",root);
Leaves(N,root);
system("pause");
return 0;
}
이 내용에 흥미가 있습니까?
현재 기사가 여러분의 문제를 해결하지 못하는 경우 AI 엔진은 머신러닝 분석(스마트 모델이 방금 만들어져 부정확한 경우가 있을 수 있음)을 통해 가장 유사한 기사를 추천합니다:
[pta] 03 - 나무 2 List Leaves (25 점)03 - 나무 2 List Leaves (25 점) Given a tree, you are supposed to list all the leaves in the order of top down, and left to...
텍스트를 자유롭게 공유하거나 복사할 수 있습니다.하지만 이 문서의 URL은 참조 URL로 남겨 두십시오.
CC BY-SA 2.5, CC BY-SA 3.0 및 CC BY-SA 4.0에 따라 라이센스가 부여됩니다.