从其给定的层级顺序遍历| Set-2
构造 BST
原文:https://www . geesforgeks . org/construct-BST-from-its-level-order-遍历-set-2/
根据给定的级别顺序遍历来构造 BST(二叉查找树)。
示例:
Input : {7, 4, 12, 3, 6, 8, 1, 5, 10}
Output :
BST:
7
/ \
4 12
/ \ /
3 6 8
/ / \
1 5 10
方法: 想法是制作一个结构元素 NodeDetails，它包含一个指向祖先的节点、最小数据和最大数据的指针。现在执行以下步骤:
- 将根节点推入节点详细信息类型的队列。
- 从队列中提取节点的详细信息，并将其与最小值和最大值进行比较。
- 检查 arr[]中是否有更多的元素，arr[i]是否可以保留为“temp.ptr”的子元素。
- 检查 arr[]中是否有更多的元素，arr[i]是否可以是“temp.ptr”的右子元素。
- 当队列变空时结束循环。
下面是上述方法的实现:
C++
// C++ program to construct BST
// using level order traversal
#include <bits/stdc++.h>
using namespace std;
// Node structure of a binary tree
struct Node {
int data;
Node* right;
Node* left;
Node(int x)
{
data = x;
right = NULL;
left = NULL;
}
};
// Structure formed to store the
// details of the ancestor
struct NodeDetails {
Node* ptr;
int min, max;
};
// Function for the preorder traversal
void preorderTraversal(Node* root)
{
if (!root)
return;
cout << root->data << " ";
// Traversing left child
preorderTraversal(root->left);
// Traversing right child
preorderTraversal(root->right);
}
// Function to make a new node
// and return its pointer
Node* getNode(int data)
{
Node* temp = new Node(0);
temp->data = data;
temp->left = NULL;
temp->right = NULL;
return temp;
}
// Function to construct the BST
Node* constructBst(int arr[], int n)
{
if (n == 0)
return NULL;
Node* root;
queue<NodeDetails> q;
// index variable to
// access array elements
int i = 0;
// Node details for the
// root of the BST
NodeDetails newNode;
newNode.ptr = getNode(arr[i++]);
newNode.min = INT_MIN;
newNode.max = INT_MAX;
q.push(newNode);
// Getting the root of the BST
root = newNode.ptr;
// Until there are no more
// elements in arr[]
while (i != n) {
// Extracting NodeDetails of a
// node from the queue
NodeDetails temp = q.front();
q.pop();
// Check whether there are more elements
// in the arr[] and arr[i] can be
// left child of 'temp.ptr' or not
if (i < n
&& (arr[i] < temp.ptr->data
&& arr[i] > temp.min)) {
// Create NodeDetails for newNode
// and add it to the queue
newNode.ptr = getNode(arr[i++]);
newNode.min = temp.min;
newNode.max = temp.ptr->data;
q.push(newNode);
// Make this 'newNode' as left child
// of 'temp.ptr'
temp.ptr->left = newNode.ptr;
}
// Check whether there are more elements
// in the arr[] and arr[i] can be
// right child of 'temp.ptr' or not
if (i < n
&& (arr[i] > temp.ptr->data
&& arr[i] < temp.max)) {
// Create NodeDetails for newNode
// and add it to the queue
newNode.ptr = getNode(arr[i++]);
newNode.min = temp.ptr->data;
newNode.max = temp.max;
q.push(newNode);
// Make this 'newNode' as right
// child of 'temp.ptr'
temp.ptr->right = newNode.ptr;
}
}
// Root of the required BST
return root;
}
// Driver code
int main()
{
int n = 9;
int arr[n] = { 7, 4, 12, 3, 6, 8, 1, 5, 10 };
// Function Call
Node* root = constructBst(arr, n);
preorderTraversal(root);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// JAVA program to construct BST
// using level order traversal
import java.io.*;
import java.util.*;
// Node class of a binary tree
class Node {
int data;
Node left, right;
public Node(int data)
{
this.data = data;
left = right = null;
}
}
// Class formed to store the
// details of the ancestors
class NodeDetails {
Node node;
int min, max;
public NodeDetails(Node node, int min, int max)
{
this.node = node;
this.min = min;
this.max = max;
}
}
class GFG {
// Function for the preorder traversal
public static void preorder(Node root)
{
if (root == null)
return;
System.out.print(root.data + " ");
// traversing left child
preorder(root.left);
// traversing right child
preorder(root.right);
}
// Function to construct BST
public static Node constructBST(int[] arr, int n)
{
// creating root of the BST
Node root = new Node(arr[0]);
Queue<NodeDetails> q = new LinkedList<>();
// node details for the root of the BST
q.add(new NodeDetails(root, Integer.MIN_VALUE,
Integer.MAX_VALUE));
// index variable to access array elements
int i = 1;
// until queue is not empty
while (!q.isEmpty()) {
// extracting NodeDetails of a node from the
// queue
NodeDetails temp = q.poll();
Node c = temp.node;
int min = temp.min, max = temp.max;
// checking whether there are more elements in
// the array and arr[i] can be left child of
// 'temp.node' or not
if (i < n && min < arr[i] && arr[i] < c.data) {
// make this node as left child of
// 'temp.node'
c.left = new Node(arr[i]);
i++;
// create new node details and add it to
// queue
q.add(new NodeDetails(c.left, min, c.data));
}
// checking whether there are more elements in
// the array and arr[i] can be right child of
// 'temp.node' or not
if (i < n && c.data < arr[i] && arr[i] < max) {
// make this node as right child of
// 'temp.node'
c.right = new Node(arr[i]);
i++;
// create new node details and add it to
// queue
q.add(
new NodeDetails(c.right, c.data, max));
}
}
// root of the required BST
return root;
}
// Driver code
public static void main(String[] args)
{
int n = 9;
int[] arr = { 7, 4, 12, 3, 6, 8, 1, 5, 10 };
// Function Call
Node root = constructBST(arr, n);
preorder(root);
}
}
C
// C# program to construct BST
// using level order traversal
using System;
using System.Collections.Generic;
// Node class of a binary tree
public class Node
{
public int data;
public Node left, right;
public Node(int data)
{
this.data = data;
left = right = null;
}
}
// Class formed to store the
// details of the ancestors
public class NodeDetails
{
public Node node;
public int min, max;
public NodeDetails(Node node, int min,
int max)
{
this.node = node;
this.min = min;
this.max = max;
}
}
class GFG{
// Function for the preorder traversal
public static void preorder(Node root)
{
if (root == null)
return;
Console.Write(root.data + " ");
// Traversing left child
preorder(root.left);
// Traversing right child
preorder(root.right);
}
// Function to construct BST
public static Node constructBST(int[] arr, int n)
{
// Creating root of the BST
Node root = new Node(arr[0]);
Queue<NodeDetails> q = new Queue<NodeDetails>();
// Node details for the root of the BST
q.Enqueue(new NodeDetails(root, int.MinValue,
int.MaxValue));
// Index variable to access array elements
int i = 1;
// Until queue is not empty
while (q.Count != 0)
{
// Extracting NodeDetails of a node
// from the queue
NodeDetails temp = q.Dequeue();
Node c = temp.node;
int min = temp.min, max = temp.max;
// Checking whether there are more
// elements in the array and arr[i]
// can be left child of
// 'temp.node' or not
if (i < n && min < arr[i] &&
arr[i] < c.data)
{
// Make this node as left child of
// 'temp.node'
c.left = new Node(arr[i]);
i++;
// Create new node details and add
// it to queue
q.Enqueue(new NodeDetails(c.left, min,
c.data));
}
// Checking whether there are more elements in
// the array and arr[i] can be right child of
// 'temp.node' or not
if (i < n && c.data < arr[i] && arr[i] < max)
{
// Make this node as right child of
// 'temp.node'
c.right = new Node(arr[i]);
i++;
// Create new node details and add it to
// queue
q.Enqueue( new NodeDetails(c.right,
c.data, max));
}
}
// Root of the required BST
return root;
}
// Driver code
public static void Main(String[] args)
{
int n = 9;
int[] arr = { 7, 4, 12, 3, 6, 8, 1, 5, 10 };
// Function Call
Node root = constructBST(arr, n);
preorder(root);
}
}
// This code is contributed by Princi Singh
java 描述语言
<script>
// Javascript program to construct BST
// using level order traversal
// Node class of a binary tree
class Node
{
constructor(data)
{
this.data = data;
this.left = null;
this.right = null;
}
}
// Class formed to store the
// details of the ancestors
class NodeDetails
{
constructor(node, min, max)
{
this.node = node;
this.min = min;
this.max = max;
}
}
// Function for the preorder traversal
function preorder(root)
{
if (root == null)
return;
document.write(root.data + " ");
// Traversing left child
preorder(root.left);
// Traversing right child
preorder(root.right);
}
// Function to construct BST
function constructBST(arr, n)
{
// Creating root of the BST
var root = new Node(arr[0])
var q = [];
// Node details for the root of the BST
q.push(new NodeDetails(root, -1000000000,
1000000000));
// Index variable to access array elements
var i = 1;
// Until queue is not empty
while (q.length != 0)
{
// Extracting NodeDetails of a node
// from the queue
var temp = q.shift();
var c = temp.node;
var min = temp.min, max = temp.max;
// Checking whether there are more
// elements in the array and arr[i]
// can be left child of
// 'temp.node' or not
if (i < n && min < arr[i] &&
arr[i] < c.data)
{
// Make this node as left child of
// 'temp.node'
c.left = new Node(arr[i]);
i++;
// Create new node details and add
// it to queue
q.push(new NodeDetails(c.left, min,
c.data));
}
// Checking whether there are more elements in
// the array and arr[i] can be right child of
// 'temp.node' or not
if (i < n && c.data < arr[i] && arr[i] < max)
{
// Make this node as right child of
// 'temp.node'
c.right = new Node(arr[i]);
i++;
// Create new node details and add it to
// queue
q.push( new NodeDetails(c.right,
c.data, max));
}
}
// Root of the required BST
return root;
}
// Driver code
var n = 9;
var arr = [ 7, 4, 12, 3, 6, 8, 1, 5, 10 ];
// Function Call
var root = constructBST(arr, n);
preorder(root);
// This code is contributed by noob2000
</script>
Output:
7 4 3 1 6 5 12 8 10
时间复杂度: O(N)
递归方法:
下面是构造 BST 的递归方法。
C++
// C++ implementation to construct a BST
// from its level order traversal
#include<bits/stdc++.h>
using namespace std;
// Node* of a BST
struct Node
{
int data;
Node* left;
Node* right;
Node(int data)
{
this->data = data;
this->left = NULL;
this->right = NULL;
}
};
Node* root;
// Function to print the inorder traversal
void preorderTraversal(Node* root)
{
if (root == NULL)
return;
cout<<(root->data) << " ";
preorderTraversal(root->left);
preorderTraversal(root->right);
}
// Function to get a new Node*
Node* getNode(int data)
{
// Allocate memory
Node* node = new Node(data);
return node;
}
// Function to cona BST from
// its level order traversal
Node* LevelOrder(Node* root, int data)
{
if (root == NULL)
{
root = getNode(data);
return root;
}
if (data <= root->data)
root->left = LevelOrder(root->left, data);
else
root->right = LevelOrder(root->right, data);
return root;
}
Node* constructBst(int arr[], int n)
{
if (n == 0)
return NULL;
root = NULL;
for(int i = 0; i < n; i++)
root = LevelOrder(root, arr[i]);
return root;
}
// Driver code
int main()
{
int arr[] = { 7, 4, 12, 3, 6,
8, 1, 5, 10 };
int n = sizeof(arr)/sizeof(arr[0]);
// Function Call
root = constructBst(arr, n);
preorderTraversal(root);
return 0;
}
// This code is contributed by pratham76
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation to construct a BST
// from its level order traversal
class GFG{
// Node of a BST
static class Node
{
int data;
Node left;
Node right;
Node(int data)
{
this.data = data;
this.left = null;
this.right = null;
}
};
static Node root;
// Function to print the inorder traversal
static void preorderTraversal(Node root)
{
if (root == null)
return;
System.out.print(root.data + " ");
preorderTraversal(root.left);
preorderTraversal(root.right);
}
// Function to get a new node
static Node getNode(int data)
{
// Allocate memory
Node node = new Node(data);
return node;
}
// Function to cona BST from
// its level order traversal
static Node LevelOrder(Node root, int data)
{
if (root == null)
{
root = getNode(data);
return root;
}
if (data <= root.data)
root.left = LevelOrder(root.left, data);
else
root.right = LevelOrder(root.right, data);
return root;
}
static Node constructBst(int []arr, int n)
{
if (n == 0)
return null;
root = null;
for(int i = 0; i < n; i++)
root = LevelOrder(root, arr[i]);
return root;
}
// Driver code
public static void main(String[] args)
{
int[] arr = { 7, 4, 12, 3, 6,
8, 1, 5, 10 };
int n = arr.length;
// Function Call
root = constructBst(arr, n);
preorderTraversal(root);
}
}
// This code is contributed by shikhasingrajput
Python 3
# Python3 implementation to construct a BST
# from its level order traversal
import math
# node of a BST
class Node:
def __init__(self, data):
self.data = data
self.left = None
self.right = None
# function to print the inorder traversal
def preorderTraversal(root):
if (root == None):
return None
print(root.data, end=" ")
preorderTraversal(root.left)
preorderTraversal(root.right)
# function to get a new node
def getNode(data):
# Allocate memory
newNode = Node(data)
# put in the data
newNode.data = data
newNode.left = None
newNode.right = None
return newNode
# function to construct a BST from
# its level order traversal
def LevelOrder(root, data):
if(root == None):
root = getNode(data)
return root
if(data <= root.data):
root.left = LevelOrder(root.left, data)
else:
root.right = LevelOrder(root.right, data)
return root
def constructBst(arr, n):
if(n == 0):
return None
root = None
for i in range(0, n):
root = LevelOrder(root, arr[i])
return root
# Driver code
if __name__ == '__main__':
arr = [7, 4, 12, 3, 6, 8, 1, 5, 10]
n = len(arr)
# Function Call
root = constructBst(arr, n)
root = preorderTraversal(root)
# This code is contributed by Srathore
C
// C# implementation to construct a BST
// from its level order traversal
using System;
class GFG{
// Node of a BST
public class Node
{
public int data;
public Node left;
public Node right;
public Node(int data)
{
this.data = data;
this.left = null;
this.right = null;
}
};
static Node root;
// Function to print the inorder traversal
static void preorderTraversal(Node root)
{
if (root == null)
return;
Console.Write(root.data + " ");
preorderTraversal(root.left);
preorderTraversal(root.right);
}
// Function to get a new node
static Node getNode(int data)
{
// Allocate memory
Node node = new Node(data);
return node;
}
// Function to cona BST from
// its level order traversal
static Node LevelOrder(Node root, int data)
{
if (root == null)
{
root = getNode(data);
return root;
}
if (data <= root.data)
root.left = LevelOrder(root.left, data);
else
root.right = LevelOrder(root.right, data);
return root;
}
static Node constructBst(int []arr, int n)
{
if (n == 0)
return null;
root = null;
for(int i = 0; i < n; i++)
root = LevelOrder(root, arr[i]);
return root;
}
// Driver code
public static void Main(string[] args)
{
int[] arr = { 7, 4, 12, 3, 6,
8, 1, 5, 10 };
int n = arr.Length;
// Function Call
root = constructBst(arr, n);
preorderTraversal(root);
}
}
// This code is contributed by rutvik_56
Output:
7 4 3 1 6 5 12 8 10
Python 代码的时间复杂度 O(N log(N))
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