C语言 03-树2 List Leaves

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

Show me the Code:

#include <stdio.h>
#include <stdlib.h>
#ifndef NULL
#define NULL 0
#endif // NULL
#define MAXSIZE 10
#define Null -1

typedef int ElementType;
struct QNode
{
    ElementType Data;
    struct QNode *Next;
};
typedef struct QNode *Position;
typedef Position Queque;

struct TNode
{
    ElementType Data;
    int left;
    int right;
} T[MAXSIZE];

Queque CreateQueue();
int add(Queque q, ElementType e);
ElementType Delete(Queque q);
int IsEmpty(Queque q);
int BuildTree();
void LevelOrderTraversal(int root);

int main()
{
    int root;
    root = BuildTree();
    LevelOrderTraversal(root);
    return 0;
}


/*队列抽象数据结构*/
Queque CreateQueue()
{
    Queque q;
    q = (Queque)malloc(sizeof(struct QNode));
    q->Data = 0;
    q->Next = NULL;
    return q;
}

int add(Queque q, ElementType e)
{
    Position p,tmp;
    if(!q)
        q = CreateQueue();
    tmp = q;
    while(tmp->Next)
    tmp = tmp->Next;
    p = (Position)malloc(sizeof(struct QNode));
    p->Data = e;
    p->Next = NULL;
    tmp->Next = p;
    return 0;
}

ElementType Delete(Queque q)
{
    ElementType First;
    Position tmp;
    if(!q || q->Next == NULL)
        return NULL;
    else
    {
        tmp = q->Next;
        q->Next = tmp->Next;
        First = tmp->Data;
        free(tmp);
        return First;
    }
}
int IsEmpty(Queque q)
{
    return(q->Next == NULL);
}

/*二叉树静态链表数据结构*/

int BuildTree()
{
    int N;
    scanf("%d",&N);
    getchar();
    if(N)
    {
        int i,root;
        int check[N];
        char cl,cr;
        for(i=0;i<N;i++)
            check[i] = 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 = Null;
            if(cr!='-')
            {
                T[i].right = cr-'0';
                check[T[i].right] = 1;
            }
            else
                T[i].right = Null;
        }
        for(i=0;i<N;i++)
        {
            if(check[i]==0){
                root = i;
                break;
            }
        }
        return root;
    }
    else
        return Null;
}

void LevelOrderTraversal(int root)
{
    Queque q;
    q = CreateQueue();
    if(root<0)
        return;
    int tmp,flag = 1;
    add(q,root);
    while(!IsEmpty(q))
    {
        tmp = Delete(q);
        if(T[tmp].left == Null && T[tmp].right == Null)
        {
            if(flag)
            {
                printf("%d",tmp);
                flag = 0;
            }
            else
                printf(" %d",tmp);
        }
        else if(T[tmp].left == Null)
            add(q,T[tmp].right);
        else if(T[tmp].right == Null)
            add(q,T[tmp].left);
        else
        {
            add(q,T[tmp].left);
            add(q,T[tmp].right);
        }
    }
}

结果

这里写图片描述

思路与问题

问题不难，包括两个部分，建树和层次遍历找叶节点。
在树的同构中我们学习了通过静态链表建树的方法，此处我们同样通过建立一个结构体数组，将输入内容先存下来，通过结构所包含的left和right变量指向儿子节点的数组下标。
层次遍历部分，不能通过递归实现，在mooc中，何钦铭老师已经讲述了通过队列实现层次遍历的方法，此处即是用这样的思路对树进行遍历，遇到没有儿子节点的结点，就将他print出来