UVa548 树 题解

先看看题：

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。

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.

样例输入：

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.

样例输出：

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.

样例输入输出example：

输入：

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

输出：

1
3
255

分析：

后续遍历的第一个字符就是根，因此只需在中序遍历中找到它，就知道子树的中序和后序遍历了

显而易见，轻轻松松构造出二叉树

一遍递归遍历

最优解自动浮出

#include
#include
#include
#include
#include
#include
using namespace std;

const int maxv=10000+10;
int in_order[maxv],post_order[maxv],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;
}

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

ps：因为各个结点的权值各不相同且都是整数，可以直接用权值作为节点编号