Tree

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 

Output 

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

题目大意：

给定一个二叉树的中序和后序遍历，求二叉树到每个叶节点的路径和最小的那个叶节点的值。

解题思路：

先建树，后dfs，建树也就是后序的最后一个就是二叉树的当前节点的值，再在中序中找到这个值，那么左边就是左子树，右边就是又子树，再从后序中找出相应的左右子树的后序，然后划分为子问题递归求解。

解题代码：

#include <iostream>
#include <cstdio>
#include <string>
#include <sstream>
#include <vector>
using namespace std;

struct node{
    int c;
    node *l,*r;
    node(){l=NULL;r=NULL;}
};

vector <int> v;

node* build(vector <int> vIn,vector <int> vPost){
    int s=0;
    node *tree=new node();
    vector <int> lIn,rIn;
    vector <int> lPost,rPost;
    tree->c=vPost.back();
    while(vIn[s]!=tree->c){
        lIn.push_back(vIn[s]);
        lPost.push_back(vPost[s]);
        s++;
    }
    if(lIn.size()>0) tree->l=build(lIn,lPost);
    for(int i=s+1;i<vIn.size();i++){
        rIn.push_back(vIn[i]);
        rPost.push_back(vPost[i-1]);
    }
    if(rIn.size()>0) tree->r=build(rIn,rPost);
    return tree;
}

void dfs(node *p,int sum){
    if(p->l==NULL && p->r==NULL ){
        v.push_back(sum+(p->c));
        v.push_back(p->c);
    }
    if(p->l!=NULL) dfs(p->l,sum+(p->c));
    if(p->r!=NULL) dfs(p->r,sum+(p->c));
}

int main(){
    string str1,str2;
    while(getline(cin,str1) &&  getline(cin,str2)){
        int x;
        vector <int> vIn,vPost;
        stringstream ss1(str1),ss2(str2);
        while(ss1>>x) vIn.push_back(x);
        while(ss2>>x) vPost.push_back(x);
        node *tree=new node();
        tree=build(vIn,vPost);
        v.clear();
        dfs(tree,0);
        int sum=1e9,ans=-1;
        for(int i=0;i<v.size();i+=2){
            if(v[i]<sum){
                sum=v[i];
                ans=v[i+1];
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}

补充：