03-树2 List Leaves (25分)

在网上看到别人的代码，真心觉得写的很好

#include<stdio.h>#include<ctype.h>struct Node{    int root;    int left;    int right;};int main(){    //create the tree    int n;    struct Node nodes[10];    scanf("%d",&n);    for(int i=0;i<n;i++){        nodes[i].root = 1;        nodes[i].left = -1;        nodes[i].right = -1;    }    for(int i=0;i<n;i++){        char ch1,ch2;        scanf("\n%c %c",&ch1,&ch2);        if(isdigit(ch1)){            nodes[i].left = ch1 - '0';            nodes[ch1-'0'].root = 0;        }        if(isdigit(ch2)){            nodes[i].right = ch2 - '0';            nodes[ch2-'0'].root = 0;        }    }    int root;    for(int i=0;i<n;i++)    {        if(nodes[i].root == 1){            root = i;            break;        }    }    //find the leave and print it;    int leaves = 0;    int queue[10];    int head=0,rear=0;    queue[rear++] = root;    while(rear-head){        int node = queue[head++];        if(nodes[node].left == -1 && nodes[node].right == -1){            if(leaves)                printf(" ");            printf("%d",node);            ++leaves;        }        if(nodes[node].left != -1){            queue[rear++] = nodes[node].left;        }        if(nodes[node].right != -1){            queue[rear++] = nodes[node].right;        }    }    return 0;}