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| Time Limit: 3000MS | Memory Limit: Unknown | 64bit IO Format: %lld & %llu |
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.
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
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;
}
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原文地址：http://www.cnblogs.com/handsomecui/p/5096983.html
