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【PAT - 甲级1020】Tree Traversals (25分)(树的遍历,给定中序后序,求层次遍历)

题干:

Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and inorder traversal sequences, you are supposed to output the level order traversal sequence of the corresponding binary tree.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (≤30), the total number of nodes in the binary tree. The second line gives the postorder sequence and the third line gives the inorder sequence. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the level order traversal sequence of the corresponding binary tree. All the numbers in a line must be separated by exactly one space, and there must be no extra space at the end of the line.

Sample Input:

7
2 3 1 5 7 6 4
1 2 3 4 5 6 7

Sample Output:

4 1 6 3 5 7 2

解题报告:

样例的树是这样的:

【PAT -  甲级1020】Tree Traversals (25分)(树的遍历,给定中序后序,求层次遍历)_#include

注意写好dfs就行了。

AC代码:

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define FF first
#define SS second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 2e5 + 5;
int n,H[MAX],Z[MAX];
struct Node {
int l,r;
int val;
} R[MAX];
int tot;
void dfs(int Zl,int Zr,int Hl,int Hr,int& root) {
root = ++tot;
int tar = H[Hr],pos,rt = root;
if(Zl > Zr) return;
R[rt].val = tar;
for(int i = Zl; i<=Zr; i++) {
if(Z[i] == tar) {pos = i;break;}
}
R[rt].val = tar;
dfs(Zl,pos-1,Hl,Hl+(pos-Zl-1),R[rt].l);
dfs(pos+1,Zr,Hl+pos-Zl,Hr-1,R[rt].r);
}
void bfs() {
int flag = 0;
queue<int> q;
q.push(1);
while(q.size()) {
int cur = q.front();q.pop();
if(flag == 1) printf(" ");
flag = 1; printf("%d",R[cur].val);
if(R[cur].l != 0 && R[R[cur].l].val != 0) q.push(R[cur].l);
if(R[cur].r != 0 && R[R[cur].r].val != 0) q.push(R[cur].r);
}
}
int main()
{
cin>>n;
for(int i = 1; i<=n; i++) cin>>H[i];
for(int i = 1; i<=n; i++) cin>>Z[i];
int xx;
dfs(1,n,1,n,xx);

bfs();
return 0 ;
}
/*
7

1 2 3 4 5 6 7


2 3 1 5 7 6 4



*/

 

错误代码1:

刚开始dfs是这么写的:

void dfs(int Zl,int Zr,int Hl,int Hr,int root) {
if(Zl >= Zr) return;
int tar = H[Hr],pos,rt = root;
R[rt].val = tar;

for(int i = Zl; i<=Zr; i++) {
if(Z[i] == tar) {pos = i;break;}
}
R[rt].val = tar;
R[rt].l = ++tot; R[rt].r = ++tot;
dfs(Zl,pos-1,Hl,Hl+(pos-Zl-1),R[rt].l);
dfs(pos+1,Zr,Hl+pos-Zl,Hr-1,R[rt].r);
}

存在问题,叶子节点根本没赋值,就return了。 

后来改成这样:

void dfs(int Zl,int Zr,int Hl,int Hr,int root) {
int tar = H[Hr],pos,rt = root;
R[rt].val = tar;
if(Zl >= Zr) return;
for(int i = Zl; i<=Zr; i++) {
if(Z[i] == tar) {pos = i;break;}
}
R[rt].val = tar;
R[rt].l = ++tot; R[rt].r = ++tot;
dfs(Zl,pos-1,Hl,Hl+(pos-Zl-1),R[rt].l);
dfs(pos+1,Zr,Hl+pos-Zl,Hr-1,R[rt].r);
}

存在问题,虽然左或右叶子节点没有值,但是还是给开了节点,也就是不确定有无值的情况下就++tot来开节点了,导致bfs中识别错误了。

 其实这样写也可以AC

AC代码2:

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define FF first
#define SS second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 2e5 + 5;
int n,H[MAX],Z[MAX];
struct Node {
int l,r;
int val;
} R[MAX];
int tot=1;
void dfs(int Zl,int Zr,int Hl,int Hr,int root) {
int tar = H[Hr],pos,rt = root;
if(Zl > Zr) return;
R[rt].val = tar;
for(int i = Zl; i<=Zr; i++) {
if(Z[i] == tar) {pos = i;break;}
}
R[rt].val = tar;
R[rt].l = ++tot; R[rt].r = ++tot;
dfs(Zl,pos-1,Hl,Hl+(pos-Zl-1),R[rt].l);
dfs(pos+1,Zr,Hl+pos-Zl,Hr-1,R[rt].r);
}
void bfs() {
int flag = 0;
queue<int> q;
q.push(1);
while(q.size()) {
int cur = q.front();q.pop();
if(flag == 1) printf(" ");
flag = 1; printf("%d",R[cur].val);
if(R[cur].l != 0 && R[R[cur].l].val != 0) q.push(R[cur].l);
if(R[cur].r != 0 && R[R[cur].r].val != 0) q.push(R[cur].r);
}
}
int main()
{
cin>>n;
for(int i = 1; i<=n; i++) cin>>H[i];
for(int i = 1; i<=n; i++) cin>>Z[i];
dfs(1,n,1,n,1);
bfs();
return 0 ;
}
/*
7

1 2 3 4 5 6 7


2 3 1 5 7 6 4



*/

总之,要让l>r 和l==r分开处理。不然就容易误杀。


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