UVA548

题意：根据中序 和 后序, 求每个叶子节点 到 根节点的最小距离

// 建立树上比较麻烦 , 但只要知道那两个递归就容易了。

// 递归的参数有注解

#include<iostream>
#include<string>
#include<sstream>
#include<algorithm>
using namespace std;

const int maxn = 10010;
int mid[maxn], post[maxn];
int linkl[maxn], linkr[maxn];
int maxxx = 1000000000, bestNode = 100000, n;

// mid(中序) 的ml, mr(左 右范围)   post(后序)的pl, pr(左右范围)
int built_mid_post(int ml, int mr, int pl, int pr){

	
	if(ml > mr) return 0;
	int root = post[pr];
	int start = ml;
	while(mid[start] != root) start++;
	int leftTreeLength = start - ml;
	linkl[root] = built_mid_post(ml, ml + leftTreeLength - 1 , pl, pl+leftTreeLength-1);
	linkr[root] = built_mid_post(ml+leftTreeLength+1, mr, pl+leftTreeLength, pr-1);
	return root;
}

void dfs(int node, int sum){

	sum += node;
	
	if(!linkl[node] && !linkr[node]) {
		if(sum < maxxx || (sum==maxxx && bestNode > node))
		{
			maxxx = sum; bestNode = node;
		}
	}
	if(linkl[node]) dfs(linkl[node], sum);
	if(linkr[node]) dfs(linkr[node], sum);
		
}

bool read_list(int *a){

	string line;
	if(!getline(cin, line)) return false;
	stringstream ss(line);
	n = 0; int  x;
	while(ss>>x ) a[n++] = x;
	return n>0;
}

int main(){



	while(read_list(mid)){

		read_list(post);
		bestNode = 100000;
		maxxx = 1000000000;
		built_mid_post(0, n-1, 0, n-1);
		dfs(post[n-1], 0);
	
	
		cout<<bestNode<<endl;
	}
	return 0;
}