Tree

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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

题意：给出一颗树的中序遍历和后序遍历；然后要你求出从根节点到叶节点所经过的结点的最小值的叶节点；若果值相等，那么输出较小的那个。

<span style="font-size:14px;"># include <cstdio>
# include <string>
# include <sstream>
# include <algorithm>
# include <iostream>
using namespace std;

int in_order[10005],out_order[10005],n,lch[10005],rch[10005],best,best_sum;

bool read_list(int *a)      // 读入字符串；
{
	string s;
	if(!getline(cin,s))    return false;   //如果为空，那么返回假，跳出循环;
	stringstream ss(s);
	n=0;
	while(ss>>a[n])   n++;     //使用流，将数据读入数组；
	return n>0;      //返回n的值;
}

int build(int l1,int r1,int l2,int r2)      //建立树；查找根结点;后序遍历的最后一个肯定是根结点；然后
{                                           //根据根结点的位置，分出根结点的左分枝和右分枝；
	if(l1>r1)    return 0;
	int root=out_order[r2],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 dfs(int root,int sum)       //从根节点开始遍历，找出和最小的叶节点;
{
	sum+=root;
	if(!lch[root]&&!rch[root])   //判断是否到叶节点;
	{
		if(sum<best_sum||(sum==best_sum&&root<best))
		{
			best_sum=sum;   best=root;
		}
	}
	if(lch[root])    dfs(lch[root],sum);
	if(rch[root])    dfs(rch[root],sum);
	return 0;
}

int main()
{
	//freopen("a.txt","r",stdin);
	while(read_list(in_order))
	{
		read_list(out_order);
		build(0,n-1,0,n-1);
		best_sum=1000000;       //赋初始值;
		dfs(out_order[n-1],0);
		cout<< best <<'\n';
	}
	return 0;
}</span>

Tree

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原文地址：http://blog.csdn.net/rechard_chen/article/details/41016017