Trie是一种有效的信息检索数据结构。使用Trie，可以使搜索复杂度达到最佳极限(密钥长度)。
给定多个字符串。任务是使用递归将字符串插入到Trie中。
例子：
Input : str = {"cat", "there", "caller", "their", "calling"}
Output : caller
calling
cat
there
their
root
/ \
c t
| |
a h
| \ |
l t e
| | \
l i r
| \ | |
e i r e
| |
r n
|
g
Input : str = {"Candy", "cat", "Caller", "calling"}
Output : caller
calling
candy
cat
root
|
c
|
a
/ | \
l n t
| |
l d
| \ |
e i y
| |
r n
|
g方法：一种有效的方法是将输入键的每个字符都视为一个单独的Trie节点，并将其插入到Trie中。请注意，子级是指向下一级Trie节点的指针(或引用)的数组。关键字符充当子数组的索引。如果输入键是新键或现有键的扩展，则需要构造键的不存在节点，并将单词的结尾标记为最后一个节点。如果输入键是Trie中现有键的前缀，我们只需将键的最后一个节点标记为单词的结尾即可。密钥长度决定了Trie深度。
下面是上述方法的实现：
CPP
#include
using namespace std;
# define CHILDREN 26
# define MAX 100
// Trie node
struct trie
{
trie *child[CHILDREN];
// endOfWord is true if the node represents
// end of a word
bool endOfWord;
};
// Function will return the new node(initialized to NULLs)
trie* createNode()
{
trie *temp = new trie();
temp->endOfWord = false;
for(int i = 0 ; i < CHILDREN ; i++)
{
// Initially , initialize null to the all child
temp->child[i] = NULL;
}
return temp;
}
// Function will insert the string in a trie recursively
void insertRecursively(trie* itr, string str, int i)
{
if(i < str.length())
{
int index = str[i] - 'a';
if(itr->child[index] == NULL )
{
// Create a new node
itr->child[index] = createNode();
}
// Recursive call for insertion of string
insertRecursively(itr->child[index], str, i + 1);
}
else
{
// Make the endOfWord true which represents
// the end of string
itr->endOfWord = true;
}
}
// Function call to insert a string
void insert(trie* itr, string str)
{
// Function call with necessary arguments
insertRecursively(itr, str, 0);
}
// Function to check whether the node is leaf or not
bool isLeafNode(trie* root)
{
return root->endOfWord != false;
}
// Function to display the content of trie
void displayContent(trie* root, char str[], int level)
{
// If node is leaf node, it indicates end
// of string, so a null character is added
// and string is displayed
if (isLeafNode(root))
{
// Assign a null character in temporary string
str[level] = '\0';
cout << str << endl;
}
for (int i = 0; i < CHILDREN; i++)
{
// If NON NULL child is found
// add parent key to str and
// call the display function recursively
// for child node
if (root->child[i])
{
str[level] = i + 'a';
displayContent(root->child[i], str, level + 1);
}
}
}
// Function call for displaying content
void display(trie* itr)
{
int level = 0;
char str[MAX];
// Function call with necessary arguments
displayContent(itr, str, level);
}
// Driver code
int main()
{
trie *root = createNode();
insert(root, "thier");
insert(root, "there");
insert(root, "answer");
insert(root, "any");
/* After inserting strings, trie will look like
root
/ \
a t
| |
n h
| \ |
s y e
| | \
w i r
| | |
e r e
|
r
*/
display(root);
return 0;
} Python3
# Python3 program to traverse in bottom up manner
CHILDREN = 26
MAX = 100
# Trie node
class trie:
def __init__(self):
self.child = [None for i in range(CHILDREN)]
# endOfWord is true if the node represents
# end of a word
self.endOfWord = False
# Function will return the new node(initialized to NULLs)
def createNode():
temp = trie()
return temp
# Function will insert the string in a trie recursively
def insertRecursively(itr, str, i):
if(i < len(str)):
index = ord(str[i]) - ord('a')
if(itr.child[index] == None ):
# Create a new node
itr.child[index] = createNode();
# Recursive call for insertion of string
insertRecursively(itr.child[index], str, i + 1);
else:
# Make the endOfWord true which represents
# the end of string
itr.endOfWord = True;
# Function call to insert a string
def insert(itr, str):
# Function call with necessary arguments
insertRecursively(itr, str, 0);
# Function to check whether the node is leaf or not
def isLeafNode(root):
return root.endOfWord != False;
# Function to display the content of trie
def displayContent(root, str, level):
# If node is leaf node, it indicates end
# of string, so a null character is added
# and string is displayed
if (isLeafNode(root)):
# Assign a null character in temporary string
print(''.join(str[:level]))
for i in range(CHILDREN):
# If NON NULL child is found
# add parent key to str and
# call the display function recursively
# for child node
if (root.child[i]):
str[level] = chr(i + ord('a'))
displayContent(root.child[i], str, level + 1);
# Function call for displaying content
def display(itr):
level = 0;
str = ['' for i in range(MAX)];
# Function call with necessary arguments
displayContent(itr, str, level);
# Driver code
if __name__=='__main__':
root = createNode();
insert(root, "thier");
insert(root, "there");
insert(root, "answer");
insert(root, "any");
''' After inserting strings, trie will look like
root
/ \
a t
| |
n h
| \ |
s y e
| | \
w i r
| | |
e r e
|
r
'''
display(root);
# This code is contributed by rutvik_56输出：
answer
any
there
their