mid-14. List Leaves （15 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 (≤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

思路： 根結點肯定不是其他節點的孩子，所以樣例中未出現的3就是根結點。構造出二叉樹，並使用佇列層序輸出。採用C++佇列操作。

#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<queue>
using namespace std;
queue<int> q;


struct TreeNode
{
	int left;
	int right;
}T[10];
int check[10];//開一個標記陣列


int Buildtree(struct TreeNode T[])
{
	int n,i,Root=-1;
	char cl,cr;
	scanf("%d",&n);
	for(i=0;i<n;i++)
	{
		scanf("\n%c %c",&cl,&cr); 
		if(cl!='-')//非空就存入
		{
			T[i].left =cl-'0';//字元轉化為數字 
			check[T[i].left ]=1;/標記已存入
		}
		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)//遇到頭節點就退出 
		break;
	}
	Root=i;//未被存入的就是頭節點 
	return Root;
}

void Output_leaves(int Root)
{
	int count=0;
	int temp;
	if(Root==-1)return;
	q.push(Root);
	while(!q.empty())
	{
	temp=q.front();//隊頭給temp
	q.pop();//出隊 
	if(T[temp].left ==-1&&T[temp].right ==-1)//若是葉子節點，則輸出
	{
		if(count++!=0)
		{
			printf(" ");//輸出空格用的 
		}
		printf("%d",temp);
	}
	if(T[temp].left !=-1)q.push(T[temp].left);//左孩子進隊 
	if(T[temp].right !=-1)q.push(T[temp].right);//右孩子進隊 
    }
}                                          


int main()
{
	int Root;
	Root=Buildtree(T);
	Output_leaves(Root);
	return 0;
}