题目
原文:
Given a binary search tree, design an algorithm which creates a linked list of all the nodes at each depth (i.e., if you have a tree with depth D, you’ll have D linked lists).
译文:
给定一棵二叉查找树,设计算法,将每一层的所有结点构建为一个链表(也就是说, 如果树有D层,那么你将构建出D个链表)
解答
这道题目本质上是个BFS,也就是说,如果已经构建了第i层结点的链表, 那么将此链表中每个结点的左右孩子结点取出,即可构建第i+1层结点的链表。 设结点类型为Node,那么指向每一层链表头结点的类型为list<Node*>, 将每一层头结点指针放到vector中。如果当前层的链表不为空, 那么将该链表的结点依次取出,然后将这些结点的不为空的孩子放入新的链表中。
代码如下:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
vector<list<Node*> > find_level_linklists(Node *head){
vector<list<Node*> > res;
int level = 0;
list<Node*> li;
li.push_back(head);
res.push_back(li);
while(!res[level].empty()){
list<Node*> l;
list<Node*>::iterator it;
for(it=res[level].begin(); it!=res[level].end(); ++it){
Node *n = *it;
if(n->lchild) l.push_back(n->lchild);
if(n->rchild) l.push_back(n->rchild);
}
++level;
res.push_back(l);
}
return res;
}
完整代码如下:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
#include <iostream>
#include <cstring>
#include <vector>
#include <list>
using namespace std;
const int maxn = 100;
struct Node{
int key;
Node *lchild, *rchild, *parent;
};
Node *p, node[maxn];
int cnt;
void init(){
p = NULL;
memset(node, '\0', sizeof(node));
cnt = 0;
}
void createMinimalTree(Node* &head, Node *parent, int a[], int start, int end){
if(start <= end){
int mid = (start + end)>>1;
node[cnt].key = a[mid];
node[cnt].parent = parent;
head = &node[cnt++];
createMinimalTree(head->lchild, head, a, start, mid-1);
createMinimalTree(head->rchild, head, a, mid+1, end);
}
}
vector<list<Node*> > find_level_linklists(Node *head){
vector<list<Node*> > res;
int level = 0;
list<Node*> li;
li.push_back(head);
res.push_back(li);
while(!res[level].empty()){
list<Node*> l;
list<Node*>::iterator it;
for(it=res[level].begin(); it!=res[level].end(); ++it){
Node *n = *it;
if(n->lchild) l.push_back(n->lchild);
if(n->rchild) l.push_back(n->rchild);
}
++level;
res.push_back(l);
}
return res;
}
void print(vector<list<Node*> > res){
vector<list<Node*> >::iterator vit;
for(vit=res.begin(); vit!=res.end(); ++vit){
list<Node*> li = *vit;
list<Node*>::iterator lit;
for(lit=li.begin(); lit!=li.end(); ++lit){
Node *n = *lit;
cout<<n->key<<" ";
}
cout<<endl;
}
}
int main(){
init();
int a[] = {
0, 1, 2, 3, 4, 5, 6, 7, 8
};
Node *head = NULL;
createMinimalTree(head, NULL, a, 0, 6);
vector<list<Node*> > res;
res = find_level_linklists(head);
print(res);
return 0;
}
全书题解目录:
Cracking the coding interview–问题与解答
全书的C++代码托管在Github上:
https://github.com/Hawstein/cracking-the-coding-interview
声明:自由转载-非商用-非衍生-保持署名 | 创意共享3.0许可证,转载请注明作者及出处
出处:http://hawstein.com/2012/12/27/4.4/