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 

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

제목:
두 갈래 나무의 후순과 중순을 정하여 이 두 갈래 나무가 뿌리 노드에서 잎 점까지의 합의 최소값을 구하고 최소값을 출력하는 잎 노드의 값 풀이 사고방식:
1 템플릿을 이용하여 두 갈래 나무 만들기
2 가장 작은 화합의 잎 노드의 값을 반복한다

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

const int maxv = 10000 + 10;

int in_order[maxv],post_order[maxv];

int lch[maxv],rch[maxv];

int n;

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;
}

// in_order[L1..R1] post_order[L2..R2] ， 
int build(int L1,int R1,int L2,int R2) {
	
	if(L1 > R1)
		return 0;
	int root = post_order[R2];
	int p = L1;
	while(in_order[p] != root)
		p++;
	int cnt = p - L1;
	
	lch[root] = build(L1 , p-1, L2 , L2+cnt-1);
	rch[root] = build(p+1, R1 , L2+cnt, R2-1);
	
	return root;
}

int best,best_sum;

void dfs(int u,int sum) {
	sum += u;
	if(!lch[u] && !rch[u]) { // 
		if(sum < best_sum || (sum == best_sum && u < best)) {
			best = u;
			best_sum = sum;
		}	
	}
	if(lch[u])
		dfs(lch[u],sum);
	if(rch[u])
		dfs(rch[u],sum);
}

int main() {
	while(read_list(in_order)) {
		read_list(post_order);
		build(0,n-1,0,n-1);
		best_sum = 0x3f3f3f;
		dfs(post_order[n-1],0);
		cout<<best<<endl;
	}
	return 0;
}

두 갈래 나무를 세우는 방법으로 이 문제를 완성하다

#include <cstdio>
#include <ctype.h>
#include <cstring>
using namespace std;
const int INF = 0x3f3f3f3f;
const int N = 100000;
struct Node {
	int val;
	Node* left;
	Node* right;
	Node() {
		val = 0;
		left = NULL;
		right = NULL;
	}
	Node(int _val) {
		val = _val;
		left = NULL;
		right = NULL;
	}
};
int post_order[N],in_order[N];
int getNum(char str[],int num[]) {
	int cnt = 0;
	for(int i = 0; str[i]; i++) {
		while(str[i] == ' ') {
			i++;
		}
		num[cnt] = 0;
		while(isdigit(str[i]) && str[i]) {
			num[cnt] = num[cnt]*10 + (str[i] - '0');
			i++;
		}
		cnt++;
		if(!str[i]) {
			break;
		}
	}
	return cnt;
}
// ，l1,r1 in_order，l2,r2 post_order
Node* build(Node* root,int l1,int r1,int l2,int r2) {
	if(l2 > r2) {
		return NULL;
	}
	root = new Node(post_order[r2]);
	int p = l1;
	while(in_order[p] != post_order[r2]) {
		p++;
	}
	int dis = p - l1;
	root->left = build(root->left,l1, l1+p-1, l2, l2+dis-1);
	root->right = build(root->right,p+1, r1, l2+dis, r2-1);
	return root;
}
int ans,max;
void dfs(Node* root,int sum) {
	sum += root->val;
	if(root->left == NULL && root->right == NULL) {
		if(sum < max || (sum == max && root->val < ans)) {
			ans = root->val;
			max = sum;
		}
	}
	if(root->left != NULL) {
		dfs(root->left,sum);
	}
	if(root->right != NULL) {
		dfs(root->right,sum);
	}
}
int main() {
	char str[N];
	while(gets(str)) {
		int n = getNum(str,in_order);
		gets(str);
		getNum(str,post_order);
		Node* root = NULL;
		root = build(root,0,n-1,0,n-1);
		max = INF;
		dfs(root,0);
		printf("%d
",ans);
	}
	return 0;
}