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最新推荐文章于 2024-03-25 15:23:31 发布

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最新推荐文章于 2024-03-25 15:23:31 发布

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博客要求根据二叉树的中序和后序遍历序列构造二叉树，然后找出从根节点到叶子节点的最短路径，并输出该路径终端叶子节点的值。输入为中序和后序遍历序列，输出为最短路径叶子节点的值。

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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

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

题意:根据中序遍历和后序遍历构造二叉树，在输出最短路径的叶子节点。

代码:

#include<iostream>
#include<string>
#include<sstream>
#include<algorithm>
using namespace std;
const int maxn=10010;

int in[maxn],post[maxn],lch[maxn],rch[maxn],n;

int read_list(int *a)
{
	string line;
	if(!getline(cin,line))
	 return 0;
	 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[r2];
    int p=l1;
    while(in[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_sum,best;
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))
	{
		read_list(post);
		build(0,n-1,0,n-1);
		best_sum=1000000000;
		dfs(post[n-1],0);
		cout<<best<<"\n";
	}
 }