【UVA 548 Tree】二叉树构造 & 遍历

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，并输出从根遍历权值最小的轮径的叶子节点，权值相同的情况下，输出叶子节点最小的那个

思路 ： 构造 & 遍历

AC代码：

#include<cstdio>
#include<cmath>
#include<deque>
#include<queue>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAX = 1e5 + 10;
const int INF = 1e9 + 7;
typedef long long LL;
int l[MAX],r[MAX],h[MAX],z[MAX],nl,ans,sum;
char c;
bool readz(){
char c;
nl = 1;
if(scanf("%d",&z[1]) == -1) return 0;
while(scanf("%c",&c) && c == ' ')
scanf("%d",&z[++nl]);
return 1;
}
void readh(){
for(int i = 1; i <= nl; i++)
scanf("%d",&h[i]);
}
int gz(int l1,int r1,int l2,int r2){
if(l1 > r1) return 0;
int t = h[r2];
int n = l1;
while(z
!= t) n++;
int nn = n - l1;
l[t] = gz(l1,n - 1,l2,l2 + nn - 1);
r[t] = gz(n + 1,r1,l2 + nn,r2 - 1);
return t;
}
void sc(int o,int vl){
vl += o;
if(!l[o] && !r[o]){
if(vl < ans || (vl == ans && o < sum)){
ans = vl;
sum = o;
return ;
}
}
if(l[o]) sc(l[o],vl);
if(r[o]) sc(r[o],vl);
}
int main()
{
while(readz()){
readh();
memset(l,0,sizeof l);
memset(r,0,sizeof r);
gz(1,nl,1,nl);
ans = sum = INF;
sc(h[nl],0);
printf("%d\n",sum);
}
return 0;
}