5-4 List Leaves (25分)

5-4 List Leaves   (25分)

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 $N$ ($\le 10$) 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:

81 -- -0 -2 7- -- -5 -4 6

Sample Output:

4 1 5

#include<stdio.h>#include<algorithm>#include<stdlib.h>#include<string.h>const int max = 11;using namespace std;struct TNode{int left;int right;}T[11];struct Node{int num;struct Node * next;};typedef struct Node * Position;struct QNode{Position front,rear;int size;};typedef struct QNode * Queue;Queue Q_create()/*创建队列*/{Queue head = (Queue)malloc(sizeof(struct QNode));head->front = NULL;head->rear = NULL;head->size = 0;return head;}bool isempty(Queue q){return (q->size == 0);}bool isfull(Queue q){return (q->size == 11);}void add(Queue q,int number){Position pnew = (Position)malloc(sizeof(struct Node));pnew->num = number;pnew->next =NULL;if(isempty(q)){q->front = q->rear = pnew;}else{q->rear->next = pnew;q->rear = q->rear->next;}q->size++;/*入队的时候要记得将队列+1*/}int del(Queue q){Position t;int num;if(isempty(q)){return NULL;}else{t = q->front;if(q->front == q->rear){q->front = q->rear = NULL;}else{q->front = q->front->next;}num = t->num;free(t);q->size--;/*出队的时候要记得将队列-1*/}return num;}int T_create()/*创建一个二叉树*/{int check[11];int n,r,i;char cl,cr;for(int i = 0 ; i< 11; i++){check[i] = 0;}scanf("%d",&n);getchar();if(n != 0){for(i = 0; i < n; i++){scanf("%c %c",&cl,&cr);getchar();if(cl != '-'){T[i].left = cl - '0';check[T[i].left] = 1;}else{T[i].left = -1;}if(cr != '-'){T[i].right = cr - '0';check[T[i].right] = 1;}else{T[i].right = -1;}}for(i = 0; i < n ; i++){if(check[i] == 0)break;}r = i;}else{return -1;}return r;}void findleaves(int t)/*采用层序遍历的方式，用队列来完成*/{Queue Q = Q_create();add(Q,t);int flag = 0;while(!isempty(Q)){t = del(Q);if(T[t].left == -1 && T[t].right == -1)/*当当前的父结点，没有左右子结点时就输出*/{if(flag == 0){printf("%d",t);flag = 1;}else{printf(" %d",t);}}if(T[t].left != -1){add(Q,T[t].left);}if(T[t].right != -1){add(Q,T[t].right);}}}int main(void){int r = T_create();findleaves(r);return 0;}