在 BST
中为给定的键指定前置和后续
原文:https://www . geesforgeks . org/in order-preventer-后继者-给定-key-bst/
最近在电商公司面试遇到一个问题。面试官问了以下问题: 有一个 BST 给出了根节点，关键部分只有整数。每个节点的结构如下:
C++
struct Node
{
int key;
struct Node *left, *right ;
};
Java 语言(一种计算机语言，尤用于创建网站)
static class Node
{
int key;
Node left, right ;
};
// This code is contributed by gauravrajput1
Python 3
class Node:
def __init__(self, key):
self.key = key
self.left = None
self.right = None
# This code is contributed by harshitkap00r
C
public class Node
{
public int key;
public Node left, right ;
};
// This code is contributed by gauravrajput1
java 描述语言
<script>
class Node {
constructor() {
this.key = 0;
this.left = null;
this.right = null;
}
}
</script>
您需要找到给定键的顺序后继和前置。如果在 BST 中找不到给定的键，则返回该键所在的两个值。
下面是达到预期结果的算法。这是一种递归方法:
Input: root node, key
output: predecessor node, successor node
1\. If root is NULL
then return
2\. if key is found then
a. If its left subtree is not null
Then predecessor will be the right most
child of left subtree or left child itself.
b. If its right subtree is not null
The successor will be the left most child
of right subtree or right child itself.
return
3\. If key is smaller then root node
set the successor as root
search recursively into left subtree
else
set the predecessor as root
search recursively into right subtree
以下是上述算法的实现:
C++
// C++ program to find predecessor and successor in a BST
#include <iostream>
using namespace std;
// BST Node
struct Node
{
int key;
struct Node *left, *right;
};
// This function finds predecessor and successor of key in BST.
// It sets pre and suc as predecessor and successor respectively
void findPreSuc(Node* root, Node*& pre, Node*& suc, int key)
{
// Base case
if (root == NULL) return ;
// If key is present at root
if (root->key == key)
{
// the maximum value in left subtree is predecessor
if (root->left != NULL)
{
Node* tmp = root->left;
while (tmp->right)
tmp = tmp->right;
pre = tmp ;
}
// the minimum value in right subtree is successor
if (root->right != NULL)
{
Node* tmp = root->right ;
while (tmp->left)
tmp = tmp->left ;
suc = tmp ;
}
return ;
}
// If key is smaller than root's key, go to left subtree
if (root->key > key)
{
suc = root ;
findPreSuc(root->left, pre, suc, key) ;
}
else // go to right subtree
{
pre = root ;
findPreSuc(root->right, pre, suc, key) ;
}
}
// A utility function to create a new BST node
Node *newNode(int item)
{
Node *temp = new Node;
temp->key = item;
temp->left = temp->right = NULL;
return temp;
}
/* A utility function to insert a new node with given key in BST */
Node* insert(Node* node, int key)
{
if (node == NULL) return newNode(key);
if (key < node->key)
node->left = insert(node->left, key);
else
node->right = insert(node->right, key);
return node;
}
// Driver program to test above function
int main()
{
int key = 65; //Key to be searched in BST
/* Let us create following BST
50
/ \
30 70
/ \ / \
20 40 60 80 */
Node *root = NULL;
root = insert(root, 50);
insert(root, 30);
insert(root, 20);
insert(root, 40);
insert(root, 70);
insert(root, 60);
insert(root, 80);
Node* pre = NULL, *suc = NULL;
findPreSuc(root, pre, suc, key);
if (pre != NULL)
cout << "Predecessor is " << pre->key << endl;
else
cout << "No Predecessor";
if (suc != NULL)
cout << "Successor is " << suc->key;
else
cout << "No Successor";
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find predecessor
// and successor in a BST
class GFG{
// BST Node
static class Node
{
int key;
Node left, right;
public Node()
{}
public Node(int key)
{
this.key = key;
this.left = this.right = null;
}
};
static Node pre = new Node(), suc = new Node();
// This function finds predecessor and
// successor of key in BST. It sets pre
// and suc as predecessor and successor
// respectively
static void findPreSuc(Node root, int key)
{
// Base case
if (root == null)
return;
// If key is present at root
if (root.key == key)
{
// The maximum value in left
// subtree is predecessor
if (root.left != null)
{
Node tmp = root.left;
while (tmp.right != null)
tmp = tmp.right;
pre = tmp;
}
// The minimum value in
// right subtree is successor
if (root.right != null)
{
Node tmp = root.right;
while (tmp.left != null)
tmp = tmp.left;
suc = tmp;
}
return;
}
// If key is smaller than
// root's key, go to left subtree
if (root.key > key)
{
suc = root;
findPreSuc(root.left, key);
}
// Go to right subtree
else
{
pre = root;
findPreSuc(root.right, key);
}
}
// A utility function to insert a
// new node with given key in BST
static Node insert(Node node, int key)
{
if (node == null)
return new Node(key);
if (key < node.key)
node.left = insert(node.left, key);
else
node.right = insert(node.right, key);
return node;
}
// Driver code
public static void main(String[] args)
{
// Key to be searched in BST
int key = 65;
/*
* Let us create following BST
* 50
* / \
* 30 70
* / \ / \
* 20 40 60 80
*/
Node root = new Node();
root = insert(root, 50);
insert(root, 30);
insert(root, 20);
insert(root, 40);
insert(root, 70);
insert(root, 60);
insert(root, 80);
findPreSuc(root, key);
if (pre != null)
System.out.println("Predecessor is " + pre.key);
else
System.out.println("No Predecessor");
if (suc != null)
System.out.println("Successor is " + suc.key);
else
System.out.println("No Successor");
}
}
// This code is contributed by sanjeev2552
计算机编程语言
# Python program to find predecessor and successor in a BST
# A BST node
class Node:
# Constructor to create a new node
def __init__(self, key):
self.key = key
self.left = None
self.right = None
# This function finds predecessor and successor of key in BST
# It sets pre and suc as predecessor and successor respectively
def findPreSuc(root, key):
# Base Case
if root is None:
return
# If key is present at root
if root.key == key:
# the maximum value in left subtree is predecessor
if root.left is not None:
tmp = root.left
while(tmp.right):
tmp = tmp.right
findPreSuc.pre = tmp
# the minimum value in right subtree is successor
if root.right is not None:
tmp = root.right
while(temp.left):
tmp = tmp.left
findPreSuc.suc = tmp
return
# If key is smaller than root's key, go to left subtree
if root.key > key :
findPreSuc.suc = root
findPreSuc(root.left, key)
else: # go to right subtree
findPreSuc.pre = root
findPreSuc(root.right, key)
# A utility function to insert a new node in with given key in BST
def insert(node , key):
if node is None:
return Node(key)
if key < node.key:
node.left = insert(node.left, key)
else:
node.right = insert(node.right, key)
return node
# Driver program to test above function
key = 65 #Key to be searched in BST
""" Let us create following BST
50
/ \
30 70
/ \ / \
20 40 60 80
"""
root = None
root = insert(root, 50)
insert(root, 30);
insert(root, 20);
insert(root, 40);
insert(root, 70);
insert(root, 60);
insert(root, 80);
# Static variables of the function findPreSuc
findPreSuc.pre = None
findPreSuc.suc = None
findPreSuc(root, key)
if findPreSuc.pre is not None:
print "Predecessor is", findPreSuc.pre.key
else:
print "No Predecessor"
if findPreSuc.suc is not None:
print "Successor is", findPreSuc.suc.key
else:
print "No Successor"
# This code is contributed by Nikhil Kumar Singh(nickzuck_007)
C
// C# program to find predecessor
// and successor in a BST
using System;
public class GFG
{
// BST Node
public
class Node
{
public
int key;
public
Node left, right;
public Node()
{}
public Node(int key)
{
this.key = key;
this.left = this.right = null;
}
};
static Node pre = new Node(), suc = new Node();
// This function finds predecessor and
// successor of key in BST. It sets pre
// and suc as predecessor and successor
// respectively
static void findPreSuc(Node root, int key)
{
// Base case
if (root == null)
return;
// If key is present at root
if (root.key == key)
{
// The maximum value in left
// subtree is predecessor
if (root.left != null)
{
Node tmp = root.left;
while (tmp.right != null)
tmp = tmp.right;
pre = tmp;
}
// The minimum value in
// right subtree is successor
if (root.right != null)
{
Node tmp = root.right;
while (tmp.left != null)
tmp = tmp.left;
suc = tmp;
}
return;
}
// If key is smaller than
// root's key, go to left subtree
if (root.key > key)
{
suc = root;
findPreSuc(root.left, key);
}
// Go to right subtree
else
{
pre = root;
findPreSuc(root.right, key);
}
}
// A utility function to insert a
// new node with given key in BST
static Node insert(Node node, int key)
{
if (node == null)
return new Node(key);
if (key < node.key)
node.left = insert(node.left, key);
else
node.right = insert(node.right, key);
return node;
}
// Driver code
public static void Main(String[] args)
{
// Key to be searched in BST
int key = 65;
/*
* Let us create following BST
* 50
* / \
* 30 70
* / \ / \
* 20 40 60 80
*/
Node root = new Node();
root = insert(root, 50);
insert(root, 30);
insert(root, 20);
insert(root, 40);
insert(root, 70);
insert(root, 60);
insert(root, 80);
findPreSuc(root, key);
if (pre != null)
Console.WriteLine("Predecessor is " + pre.key);
else
Console.WriteLine("No Predecessor");
if (suc != null)
Console.WriteLine("Successor is " + suc.key);
else
Console.WriteLine("No Successor");
}
}
// This code is contributed by aashish1995
java 描述语言
<script>
// JavaScript program to find predecessor
// and successor in a BST// BST Node
class Node
{
constructor(key)
{
this.key = key;
this.left = this.right = null;
}
}
var pre = new Node(), suc = new Node();
// This function finds predecessor and
// successor of key in BST. It sets pre
// and suc as predecessor and successor
// respectively
function findPreSuc(root , key)
{
// Base case
if (root == null)
return;
// If key is present at root
if (root.key == key)
{
// The maximum value in left
// subtree is predecessor
if (root.left != null)
{
var tmp = root.left;
while (tmp.right != null)
tmp = tmp.right;
pre = tmp;
}
// The minimum value in
// right subtree is successor
if (root.right != null)
{
var tmp = root.right;
while (tmp.left != null)
tmp = tmp.left;
suc = tmp;
}
return;
}
// If key is smaller than
// root's key, go to left subtree
if (root.key > key)
{
suc = root;
findPreSuc(root.left, key);
}
// Go to right subtree
else
{
pre = root;
findPreSuc(root.right, key);
}
}
// A utility function to insert a
// new node with given key in BST
function insert(node , key)
{
if (node == null)
return new Node(key);
if (key < node.key)
node.left = insert(node.left, key);
else
node.right = insert(node.right, key);
return node;
}
// Driver code
// Key to be searched in BST
var key = 65;
/*
* Let us create following BST
* 50
* / \
* 30 70
* / \ / \
* 20 40 60 80
*/
var root = new Node();
root = insert(root, 50);
insert(root, 30);
insert(root, 20);
insert(root, 40);
insert(root, 70);
insert(root, 60);
insert(root, 80);
findPreSuc(root, key);
if (pre != null)
document.write("Predecessor is " + pre.key);
else
document.write("No Predecessor");
if (suc != null)
document.write("<br/>Successor is " + suc.key);
else
document.write("<br/>No Successor");
// This code contributed by gauravrajput1
</script>
输出:
Predecessor is 60
Successor is 70
另一种方法: 我们也可以使用有序遍历来找到有序的后继者和有序的前驱者。检查当前节点是否小于前一个和后一个的给定键，检查它是否大于给定键。如果它大于给定的键，则检查它是否小于后继中已经存储的值，然后更新它。最后，获取存储在 q(后继)和 p(前驱)中的前驱和后继。
C++
// CPP code for inorder successor
// and predecessor of tree
#include<iostream>
#include<stdlib.h>
using namespace std;
struct Node
{
int data;
Node* left,*right;
};
// Function to return data
Node* getnode(int info)
{
Node* p = (Node*)malloc(sizeof(Node));
p->data = info;
p->right = NULL;
p->left = NULL;
return p;
}
/*
since inorder traversal results in
ascending order visit to node , we
can store the values of the largest
no which is smaller than a (predecessor)
and smallest no which is large than
a (successor) using inorder traversal
*/
void find_p_s(Node* root,int a,
Node** p, Node** q)
{
// If root is null return
if(!root)
return ;
// traverse the left subtree
find_p_s(root->left, a, p, q);
// root data is greater than a
if(root&&root->data > a)
{
// q stores the node whose data is greater
// than a and is smaller than the previously
// stored data in *q which is successor
if((!*q) || (*q) && (*q)->data > root->data)
*q = root;
}
// if the root data is smaller than
// store it in p which is predecessor
else if(root && root->data < a)
{
*p = root;
}
// traverse the right subtree
find_p_s(root->right, a, p, q);
}
// Driver code
int main()
{
Node* root1 = getnode(50);
root1->left = getnode(20);
root1->right = getnode(60);
root1->left->left = getnode(10);
root1->left->right = getnode(30);
root1->right->left = getnode(55);
root1->right->right = getnode(70);
Node* p = NULL, *q = NULL;
find_p_s(root1, 55, &p, &q);
if(p)
cout << p->data;
if(q)
cout << " " << q->data;
return 0;
}
Python 3
""" Python3 code for inorder successor
and predecessor of tree """
# A Binary Tree Node
# Utility function to create a new tree node
class getnode:
# Constructor to create a new node
def __init__(self, data):
self.data = data
self.left = None
self.right = None
"""
since inorder traversal results in
ascendingorder visit to node , we
can store the values of the largest
o which is smaller than a (predecessor)
and smallest no which is large than
a (successor) using inorder traversal
"""
def find_p_s(root, a, p, q):
# If root is None return
if(not root):
return
# traverse the left subtree
find_p_s(root.left, a, p, q)
# root data is greater than a
if(root and root.data > a):
# q stores the node whose data is greater
# than a and is smaller than the previously
# stored data in *q which is successor
if((not q[0]) or q[0] and
q[0].data > root.data):
q[0] = root
# if the root data is smaller than
# store it in p which is predecessor
elif(root and root.data < a):
p[0]= root
# traverse the right subtree
find_p_s(root.right, a, p, q)
# Driver Code
if __name__ == '__main__':
root1 = getnode(50)
root1.left = getnode(20)
root1.right = getnode(60)
root1.left.left = getnode(10)
root1.left.right = getnode(30)
root1.right.left = getnode(55)
root1.right.right = getnode(70)
p = [None]
q = [None]
find_p_s(root1, 55, p, q)
if(p[0]) :
print(p[0].data, end = "")
if(q[0]) :
print("", q[0].data)
# This code is contributed by
# SHUBHAMSINGH10
java 描述语言
<script>
class Node
{
constructor(data)
{
this.data = data;
this.left = this.right = null;
}
}
function find_p_s(root, a, p, q)
{
// If root is None return
if(root == null)
return
// traverse the left subtree
find_p_s(root.left, a, p, q)
// root data is greater than a
if(root && root.data > a)
{
// q stores the node whose data is greater
// than a and is smaller than the previously
// stored data in *q which is successor
if((q[0] == null) || q[0] != null && q[0].data > root.data)
q[0] = root
}
// if the root data is smaller than
// store it in p which is predecessor
else if(root && root.data < a)
{ p[0] = root
}
// traverse the right subtree
find_p_s(root.right, a, p, q)
}
// Driver Code
let root1 = new Node(50)
root1.left = new Node(20)
root1.right = new Node(60)
root1.left.left = new Node(10)
root1.left.right = new Node(30)
root1.right.left = new Node(55)
root1.right.right = new Node(70)
p = [null]
q = [null]
find_p_s(root1, 55, p, q)
if(p[0] != null)
document.write(p[0].data, end = " ")
if(q[0] != null)
document.write(" ", q[0].data)
// This code is contributed by patel2127
</script>
输出:
50 60
感谢 Shweta 提出这个方法。
？list = plqm 7 alhxfyshcxd 7r 1j0ky 9 ZG _ gbb 1 dbk〖t0〗]
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