UVA-548Tree（二叉树的递归遍历）

Tree

Time Limit: 3000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

Submit Status

Description

 

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

题解：给出了一个二叉树的中序和后序遍历，让求树枝的最小总长的最小枝节点；思路是把这个二叉树求出来，然后找最小值，找最小值的地方错了好多次；必须要先找最小总长，再找相等的情况下最小枝节点；另外后序遍历其实倒过来就是先序遍历；在后序遍历的最后一个点其实就是根节点，根据这个在中序遍历里面递归找左右树；

代码：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
#define SI(x) scanf("%d",&x)
#define mem(x,y) memset(x,y,sizeof(x))
#define PI(x) printf("%d",x)
#define P_ printf(" ")
const int INF=0x3f3f3f3f;
typedef long long LL;
const int MAXN=100010;
int v1[MAXN],v2[MAXN];
char s[MAXN];
int ans,minlength;
struct Node{
	int v;
	Node *L,*R;
	Node(){
		L=NULL;R=NULL;
	}
};
int init(char *s,int *v){
	int k=0;
	for(int i=0;s[i];i++){
		while(isdigit(s[i]))v[k]=v[k]*10+s[i++]-'0';
		k++;
	}
//	for(int i=0;i<k;i++)printf("%d ",v[i]);puts("");
	return k;
}
int find(int n,int *v,int t){
	for(int i=n-1;i>=0;i--)
		if(v[i]==t)return i;
		 return 0;
}
Node* build(int n,int *v1,int *v2){
	Node *root;
	if(n<=0)return NULL;
	root=new Node;
	root->v=v2[n-1];
	int p=find(n,v1,v2[n-1]);
	root->L=build(p,v1,v2);
	root->R=build(n-p-1,v1+p+1,v2+p);
	return root;
}
void dfs(Node *root,int length){
	if(root==NULL)return;
	length+=root->v;
	if(root->L==NULL&&root->R==NULL){
		if(minlength>length)
        {
            minlength=length;
            ans=root->v;
        }
        else if(minlength==length)
        ans=min(ans,root->v);
		return;
	}
	dfs(root->L,length);
	dfs(root->R,length);
}
int main(){
	while(gets(s)){
		mem(v1,0);mem(v2,0);
		init(s,v1);
		gets(s);
		int k=init(s,v2);
		Node *root=build(k,v1,v2);
		ans=minlength=INF;
		dfs(root,0);
		printf("%d\n",ans);
	}
	return 0;
}