UVA548 Tree （二叉树的遍历）

You are to determine the value of the leaf node in a given binary tree that is the terminal node of a

path of least value from the root of the binary tree to any leaf. The value of a path is the sum of values

of nodes along that path.

Input

The input file will contain a description of the binary tree given as the inorder and postorder traversal

sequences of that tree. Your program will read two line (until end of file) from the input file. The first

line will contain the sequence of values associated with an inorder traversal of the tree and the second

line will contain the sequence of values associated with a postorder traversal of the tree. All values

will be different, greater than zero and less than 10000. You may assume that no binary tree will have

more than 10000 nodes or less than 1 node.

Output

For each tree description you should output the value of the leaf node of a path of least value. In the

case of multiple paths of least value you should pick the one with the least value on the terminal node.

Sample Input

3 2 1 4 5 7 6

3 1 2 5 6 7 4

7 8 11 3 5 16 12 18

8 3 11 7 16 18 12 5

255

255

Sample Output

1

3

255

【分析】嗯  开始学习二叉树了，感觉很难，这个代码不是很懂，感觉像是先通过中序遍历，后序遍历求出中序遍历，找到子叶点求和。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#include<functional>
#define mod 1000000007
#define inf 0x3f3f3f3f
#define pi acos(-1.0)
using namespace std;
typedef long long ll;
const int N=300;
const int M=15005;
char s[100005];
int v1[100005],v2[100005],top;
int min_sum,ans;
int init(char *s,int *v) {
    int top=0;
    for(int i=0; s[i]; i++) {
        while(s[i]==‘ ‘)
            i++;
        v[top]=0;
        while(s[i]&&isdigit(s[i])) {
            v[top]=v[top]*10+s[i]-‘0‘;
            i++;
        }
        top++;
        if(!s[i]) break;
    }
    return top;
}
int find(int *v,int n,int c) {
    for(int i=n-1; i>=0; i--)
        if(v[i]==c)
            return i;
    return 0;
}
void build(int n,int *v1,int *v2,int sum) {
    if(n<=0)
        return ;
    int p=find(v1,n,v2[n-1]);
    //printf("v2[n-1]=%d p=%d\n",v2[n-1],p);
    sum+=v2[n-1];
    if(p<=0&&n-p-1<=0) {
        if(sum==min_sum)
            ans=min(ans,v2[n-1]);
        else if(sum<min_sum) {
            min_sum=sum;
            ans=v2[n-1];
        }
        return ;
    }
    build(p,v1,v2,sum);
    build(n-p-1,v1+p+1,v2+p,sum);
}

int main() {
    while(gets(s)) {
        int v;
        init(s,v1);
        gets(s);
        top=init(s,v2);
        ans=min_sum=10000000;
        build(top,v1,v2,0);
        printf("%d\n",ans);
    }
    return 0;
}