UVA 548 Tree(数据结构，二叉树，遍历)

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=104&page=show_problem&problem=489

  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 63 1 2 5 6 7 47 8 11 3 5 16 12 188 3 11 7 16 18 12 5255255

Sample Output 

13255

---

Miguel A. Revilla 
1999-01-11

给出二叉树的中根遍历和后根遍历，找出根节点到叶子节点权值和最小的路径，输出该叶子节点

建树+DFS

建树的过程完全考察对二叉树遍历及递归的理解，下面给出递归定义的二叉树T的先/中/后根遍历：

先根：T的根节点+T的左子树+T的右子树

中根：T的左子树+T的根节点+T的右子树

后根：T的左子树+T的右子树+T的根节点

已知中根和另一个遍历就可确定一棵树（确定根节点，分出左右子树）

多手动模拟几次建树就能明白过程了，对基本功要求比较高

DFS可以加个剪枝，不过似乎没什么效果

#include<cstdio>#include<iostream>#include<cstdlib>#include<algorithm>#include<ctime>#include<cctype>#include<cmath>#include<string>#include<cstring>#include<stack>#include<queue>#include<list>#include<vector>#include<map>#include<set>#define sqr(x) ((x)*(x))#define LL long long#define INF 0x3f3f3f3f#define PI 3.1415926535897932384626#define eps 1e-10#define mm 10001using namespace std;int InOrder[mm],PostOrder[mm];struct Node{    int v,l,r;}node[mm];int n,cnt,MIN,ANS;int build(int Il,int Ir,int Pl,int Pr){    //printf("Il=%d Ir=%d Pl=%d Pr=%d:\n",Il,Ir,Pl,Pr);    int k;    if (Il==Ir)   return -1;    for (int i=Il;i<Ir;i++)    {        if (InOrder[i]==PostOrder[Pr-1])        {            k=i;            break;        }    }    int t=cnt++;    node[t].v=InOrder[k];    node[t].l=build(Il,k,Pl,Pl+k-Il);    node[t].r=build(k+1,Ir,Pr-Ir+k,Pr-1);    return t;}void dfs(int x,int s){    if (s>MIN)  return ;    if (node[x].l==-1 && node[x].r==-1)    {        if (s+node[x].v<MIN)        {            MIN=s+node[x].v;            ANS=node[x].v;            return ;        }    }    if (node[x].l!=-1)  dfs(node[x].l,s+node[x].v);    if (node[x].r!=-1)  dfs(node[x].r,s+node[x].v);}int main(){    #ifndef ONLINE_JUDGE        freopen("/home/fcbruce/文档/code/t","r",stdin);    #endif // ONLINE_JUDGE    char c;    while (scanf("%d%c",InOrder+0,&c)==2)    {        n=1;        while (true)        {            if (c=='\n')    break;            scanf("%d%c",InOrder+n,&c);            n++;        }        //printf("%d\n",n);        for (int i=0;i<n;i++)            scanf("%d%c",PostOrder+i,&c);        cnt=0;        build(0,n,0,n);        MIN=INF;        dfs(0,0);        printf("%d\n",ANS);    }    return 0;}