Tree UVA - 548 （中序和后序，求根节点到叶节点的和最小）

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最新推荐文章于 2022-07-04 23:04:59 发布

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分类专栏：二叉树SBT

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给定一棵二叉树的中序和后序遍历序列，你需要找到从根节点到叶节点的路径，使得路径上的节点值之和最小。如果有多个这样的路径，输出叶节点值最小的那个。输入包含多组数据，每组数据对应一棵树的中序和后序遍历序列，所有数值都在1到10000之间，且不同。输出是每个树的最小路径和对应的叶节点值。

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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

题目大意：已知一棵树的中序和后序，求一棵树的根节点到叶节点的权值和最小，（注意：如果到达叶节点的和相等，输出叶节点值小的），输出的是叶节点的编号。

思路：可以通过数组模拟，将树建立起来，然后通过dfs递归遍历，求出最小的值。

通过建树以及dfs：https://blog.youkuaiyun.com/qq_41890797/article/category/8240014

#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
using namespace std;
#define inf 0x3f3f3f3f
char a[80050],b[80050];
int in[10010],after[10010],lson[10010],rson[10010];//中序，后序，左儿子，右儿子
int build(int L1,int R1,int L2,int R2)
{
    if(L1>R1)
        return 0;
    int root=after[R2];
    int p=L1;
    while(in[p]!=root)
        p++;
    int k=p-L1;//左子树
    lson[root]=build(L1,p-1,L2,L2+k-1);//访问左子树
    rson[root]=build(p+1,R1,L2+k,R2-1);//访问右子树
    return root;
}
int kk,ans;
void dfs(int u,int sum)
{
    sum+=u;
    if(!lson[u]&&!rson[u])
    {
        if(sum<ans||(sum==ans&&u<kk))
        {
            kk=u;
            ans=sum;
        }
    }
    if(lson[u])
        dfs(lson[u],sum);
    if(rson[u])
        dfs(rson[u],sum);
}
int main()
{
    while(gets(a))
    {
        gets(b);
        memset(in,0,sizeof(in));
        memset(after,0,sizeof(after));
        memset(lson,0,sizeof(lson));
        memset(rson,0,sizeof(rson));


        int l=strlen(a);
        a[l]=' ';
        b[l]=' ';
        int q=0;
        int r=0;
        for(int i=0; i<=l; i++)
        {
            if(a[i]==' ')
            {
                in[r++]=q;
                q=0;
                continue;
            }
            else
                q=q*10+(a[i]-48);
        }

        r=0,q=0;
        for(int i=0; i<=l; i++)
        {
            if(b[i]==' ')
            {
                after[r++]=q;
                q=0;
                continue;
            }
            else
                q=q*10+(b[i]-48);
        }


        int n=r;
        build(0,n-1,0,n-1);
        ans=inf;
        dfs(after[n-1],0);
        printf("%d\n",kk);
    }
    return 0;
}