UVA - 548 Tree（二叉树的递归遍历）

题意：已知中序后序序列，求一个叶子到根路径上权和最小，如果多解，则叶子权值尽量小。

分析：已知中序后序建树，再dfs求从根到各叶子的权和比较大小

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
typedef long long ll;
typedef unsigned long long llu;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {, , -, };
const int dc[] = {-, , , };
const double pi = acos(-1.0);
const double eps = 1e-;
const int MAXN =  + ;
const int MAXT =  + ;
using namespace std;
string a, b;
vector<int> in_order, post_order;
int leftchild[MAXN];
int rightchild[MAXN];
int anssum;
int ansv;
int build_tree(int L1, int R1, int L2, int R2){
    if(L1 > R1) return ;
    int root = post_order[R2];
    int st = L1;
    while(in_order[st] != root) ++st;
    int cnt = st - L1;//一定要通过个数来控制取出来的中序后序序列的左右下标
    leftchild[root] = build_tree(L1, st - , L2, L2 + cnt - );//值为root的左孩子结点的值,第四个参数不能写成st-1，因为取出来的相对应的中序和后序序列不一定是下标对齐的
    rightchild[root] = build_tree(st + , R1, L2 + cnt, R2 - );
    return root;
}
void dfs(int root, int sum){
    sum += root;
    if(!leftchild[root] && !rightchild[root]){//叶子
        if(sum < anssum || (sum == anssum && root < ansv)){
            anssum = sum;
            ansv = root;
        }
    }
    if(leftchild[root]){
        dfs(leftchild[root], sum);
    }
    if(rightchild[root]){
        dfs(rightchild[root], sum);
    }
}
int main(){
    while(getline(cin, a)){
        in_order.clear();
        post_order.clear();
        memset(leftchild, , sizeof leftchild);
        memset(rightchild, , sizeof rightchild);
        stringstream s1(a);
        int x;
        while(s1 >> x){
            in_order.push_back(x);
        }
        getline(cin, b);
        stringstream s2(b);
        while(s2 >> x){
            post_order.push_back(x);
        }
        int len = in_order.size();
        build_tree(, len - , , len - );
        int root = post_order[len - ];
        anssum = INT_M_INF;
        ansv = INT_M_INF;
        dfs(root, );
        printf("%d\n", ansv);
    }
}

已知中序和后序可建树，建成后，可输出前序序列。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
typedef long long ll;
typedef unsigned long long llu;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {, , -, };
const int dc[] = {-, , , };
const double pi = acos(-1.0);
const double eps = 1e-;
const int MAXN =  + ;
const int MAXT =  + ;
using namespace std;
string a, b;
vector<int> in_order, post_order, pre_order;
int leftchild[MAXN];
int rightchild[MAXN];
int build_tree(int L1, int R1, int L2, int R2){
    if(L1 > R1) return ;
    int root = post_order[R2];
    int st = L1;
    while(in_order[st] != root) ++st;
    int cnt = st - L1;//一定要通过个数来控制取出来的中序后序序列的左右下标
    leftchild[root] = build_tree(L1, st - , L2, L2 + cnt - );//值为root的左孩子结点的值,第四个参数不能写成st-1，因为取出来的相对应的中序和后序序列不一定是下标对齐的
    rightchild[root] = build_tree(st + , R1, L2 + cnt, R2 - );
    return root;
}
void dfs(int root){
    pre_order.push_back(root);
    if(leftchild[root]){
        dfs(leftchild[root]);
    }
    if(rightchild[root]){
        dfs(rightchild[root]);
    }
}
int main(){
    while(getline(cin, a)){
        in_order.clear();
        post_order.clear();
        pre_order.clear();
        memset(leftchild, , sizeof leftchild);
        memset(rightchild, , sizeof rightchild);
        stringstream s1(a);
        int x;
        while(s1 >> x){
            in_order.push_back(x);
        }
        getline(cin, b);
        stringstream s2(b);
        while(s2 >> x){
            post_order.push_back(x);
        }
        int len = in_order.size();
        build_tree(, len - , , len - );
        int root = post_order[len - ];
        dfs(root);
        for(int i = ; i < len; ++i){
            if(i) printf(" ");
            printf("%d", pre_order[i]);
        }
        printf("\n");
    }
}