UVA699
2017-03-08 21:59
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题目链接:http://www.cnblogs.com/cute/p/3640416.html
代码:
小记:定义一个数组来记录答案。根节点对应的那条数值线节点之和存储在数组的终点。然后dfs中递归调用,递归出口是遇到空树。
代码:
#include<stdio.h> #include<string.h> #include<iostream> using namespace std; #define maxn 10000 int res[maxn]; ///这种做法通俗易懂!!! void dfs(int val,int pos) { if (val == -1) return ; res[pos] += val; int left,right; scanf("%d",&left); dfs(left,pos - 1); scanf("%d",&right); dfs(right,pos + 1); return ; } int main () { int val; int cas = 0; while (scanf("%d",&val) && val != -1) { memset(res , 0 , sizeof(res)); dfs (val , maxn/2); printf("Case %d:\n",++cas); int p=0; while(res[p]==0) p++; cout<<res[p++]; while(res[p]!=0) cout<<" "<<res[p++]; printf("\n\n"); } return 0; }
小记:定义一个数组来记录答案。根节点对应的那条数值线节点之和存储在数组的终点。然后dfs中递归调用,递归出口是遇到空树。
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