03-树2 List Leaves

最新推荐文章于 2024-09-07 15:26:33 发布

转载最新推荐文章于 2024-09-07 15:26:33 发布 · 46 阅读

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原文链接：http://www.cnblogs.com/shin0324/p/9801636.html

本文介绍了一种算法，用于按从上到下、从左到右的顺序列出给定树的所有叶子节点。通过建立树结构并使用队列进行层序遍历来实现，最后输出所有叶子节点的索引。

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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

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

 1 #include<stdio.h>
 2 #include<stdlib.h> 
 3 #define max 10
 4 #define Null -1
 5 
 6 struct TNode{
 7     int data;
 8     int left, right;
 9 }T[max];
10 
11 struct QNode{
12     int *data;
13     int front, rear;
14 };
15 typedef struct QNode *Queue;
16 
17 //建树函数，返回根结点Root 
18 int BuildTree( ){
19     int N , Root= Null, check[max]={0};
20     char L, R;
21     scanf("%d", &N);
22     for(int i=0; i<N; i++){
23         scanf("\n%c %c", &L, &R);
24         T[i].data = i;
25         if(L=='-'){
26             T[i].left = Null;
27         }
28         else {
29             T[i].left = L-'0';
30             check[T[i].left] = 1;
31         }
32         if(R=='-'){
33             T[i].right = Null;
34         }
35         else {
36             T[i].right = R-'0';
37             check[T[i].right] = 1;
38         }
39     }
40     for(int i=0; i<N; i++){
41         if(!check[i]){
42             Root = i;
43             break;
44         }
45     }
46     return Root;
47 }
48 
49 //初始化队列函数，返回队列表头 
50 Queue CreateQueue(Queue Q){
51     Q = (Queue)malloc(sizeof(struct QNode));
52     Q->data = (int*)malloc(max*sizeof(int));
53     Q->front = Q->rear = 0;
54     return Q;
55 }
56 
57 //进队列函数，因为知道数据最大个数，一定不会溢出，所以不需要判断是否队满 
58 void Add(Queue Q, int n) {
59     Q->data[Q->front++] = n;
60 }
61 
62 bool IsEmpty(Queue Q){
63     return(Q->front==Q->rear);
64 }
65 
66 //出队列函数 
67 int Delete(Queue Q){
68     int temp = Q->data[Q->rear++];
69     return temp;
70 }
71 
72 void LevelorderTraversal(int Root){
73     if(Root==Null) return;            //如果树为空直接返回 
74     int flag = 0;        //flag用于调整输出格式 
75     Queue Q = CreateQueue(Q);         
76     Add(Q, Root);       
77     while(!IsEmpty(Q)){            //只要队列不空 
78         int temp = Delete(Q);      //将队尾元素出队 
79         if(T[temp].left==Null && T[temp].right==Null){    //如果该元素的左右子树均为空，即leaves 
80             if(!flag) flag=1;
81             else printf(" ");
82             printf("%d", temp);                           //打印该元素 
83         }
84         if(T[temp].left!=Null) Add(Q, T[temp].left);      //如果该元素左子树不空，将左子树的根结点进队 
85         if(T[temp].right!=Null) Add(Q, T[temp].right);    //同理右子树  因为题意要从左到右，所以先判断左子树在判断右子树 
86     }
87 }
88 
89 //主程序框架 
90 int main( ){
91     int Root;
92     Root = BuildTree( );
93     LevelorderTraversal(Root);
94     return 0;
95 }