PAT 03-2. List Leaves (25)

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:

8

1 -

- -

0 -

2 7

- -

- -

5 -

4 6

Sample Output:
4 1 5

#include<cstdio>
#include<cstdlib>
using namespace std;
int arr[100][2];
int list[100] = { 0 };
int result[100] = { -1 };
int count = 0;
struct TreeNode//树节点
{
int data;
struct TreeNode* left;
struct TreeNode* right;
};

typedef struct TreeNode* PtrToNode;
typedef PtrToNode Tree;

Tree Build(int x, Tree &t)//根据数组建树
{

if (x == -1)
{
t = NULL;
}
else
{
t = (Tree)malloc(sizeof(TreeNode));
t->data = x;
Build(arr[x][0], t->left);
Build(arr[x][1], t->right);
}
return t;
}

struct QueueNode//队列节点
{
Tree data;
struct QueueNode* next;
};

typedef struct QueueNode* Queue;

void push(Tree d, Queue q)//入队
{
while (q->next != NULL)
{
q = q->next;
}
Queue qt = (Queue)malloc(sizeof(struct QueueNode));
qt->data = d;
q->next = qt;
qt->next = NULL;
}

void pop(Queue q)//出队
{
Queue first = q->next;
q->next = first->next;
free(first);
}

Tree top(Queue q)//查看队列第一个元素
{
return q->next->data;
}

void view(Tree bt)//层序遍历 满足左右儿子都是NULL的打印
{
QueueNode qt;
Queue q = &qt;
q->next = NULL;
Tree t;
if (!bt)
return;
push(bt, q);
while (q->next != NULL)
{
t = top(q);
pop(q);
if (t->left == NULL&&t->right == NULL)
{
result[count] = t->data;
count++;
}
if (t->left)
push(t->left, q);
if (t->right)
push(t->right, q);
}
}

int main()
{
int n;
scanf("%d", &n);
int i;
for (i = 0; i < n; i++)//将测试用例存入数组 不存在用-1表示
{
char str1[3], str2[3];
int l, r;
scanf("%s%s", str1, str2);
if (l = atoi(str1))
{
arr[i][0] = l;
list[l]++;
}
else if (str1[0] == '0')
{
l = 0;
arr[i][0] = l;
list[l]++;
}
else
{
arr[i][0] = -1;
}
if (r = atoi(str2))
{
arr[i][1] = r;
list[r]++;
}
else if (str2[0] == '0')
{
r = 0;
arr[i][1] = r;
list[r]++;
}
else
{
arr[i][1] = -1;
}
}
for (i = 0; i < n; i++)//没有出现的数字即为根节点
{
if (list[i] == 0)
break;
}
struct TreeNode t;
Tree bt;
bt = Build(i, bt);
view(bt);
for (i = 0; i < count; i++)
{
if (i == count - 1)
printf("%d\n", result[i]);
else
printf("%d ", result[i]);
}
}