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:
注意写好dfs就行了。
AC代码:
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:
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。