给定一个排序数组。编写一个函数，使用数组元素创建一个平衡的二进制搜索树。 例如：
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Input: Array {1, 2, 3}Output: A Balanced BST 2 / 1 3 Input: Array {1, 2, 3, 4}Output: A Balanced BST 3 / 2 4 /1
算法 在 以前的职位 我们讨论了从排序链表构造BST。在O（n）时间内从排序数组构造更简单，因为我们可以在O（1）时间内得到中间元素。下面是一个简单的算法，我们首先找到列表的中间节点，并使其成为要构造的树的根。
1) Get the Middle of the array and make it root.2) Recursively do same for left half and right half. a) Get the middle of left half and make it left child of the root created in step 1. b) Get the middle of right half and make it right child of the root created in step 1.
下面是上述算法的实现。创建平衡BST的主代码突出显示。
C++
// C++ program to print BST in given range #include<bits/stdc++.h> using namespace std; /* A Binary Tree node */ class TNode { public : int data; TNode* left; TNode* right; }; TNode* newNode( int data); /* A function that constructs Balanced Binary Search Tree from a sorted array */ TNode* sortedArrayToBST( int arr[], int start, int end) { /* Base Case */ if (start > end) return NULL; /* Get the middle element and make it root */ int mid = (start + end)/2; TNode *root = newNode(arr[mid]); /* Recursively construct the left subtree and make it left child of root */ root->left = sortedArrayToBST(arr, start, mid - 1); /* Recursively construct the right subtree and make it right child of root */ root->right = sortedArrayToBST(arr, mid + 1, end); return root; } /* Helper function that allocates a new node with the given data and NULL left and right pointers. */ TNode* newNode( int data) { TNode* node = new TNode(); node->data = data; node->left = NULL; node->right = NULL; return node; } /* A utility function to print preorder traversal of BST */ void preOrder(TNode* node) { if (node == NULL) return ; cout << node->data << " " ; preOrder(node->left); preOrder(node->right); } // Driver Code int main() { int arr[] = {1, 2, 3, 4, 5, 6, 7}; int n = sizeof (arr) / sizeof (arr[0]); /* Convert List to BST */ TNode *root = sortedArrayToBST(arr, 0, n-1); cout << "PreOrder Traversal of constructed BST " ; preOrder(root); return 0; } // This code is contributed by rathbhupendra |
C
#include<stdio.h> #include<stdlib.h> /* A Binary Tree node */ struct TNode { int data; struct TNode* left; struct TNode* right; }; struct TNode* newNode( int data); /* A function that constructs Balanced Binary Search Tree from a sorted array */ struct TNode* sortedArrayToBST( int arr[], int start, int end) { /* Base Case */ if (start > end) return NULL; /* Get the middle element and make it root */ int mid = (start + end)/2; struct TNode *root = newNode(arr[mid]); /* Recursively construct the left subtree and make it left child of root */ root->left = sortedArrayToBST(arr, start, mid-1); /* Recursively construct the right subtree and make it right child of root */ root->right = sortedArrayToBST(arr, mid+1, end); return root; } /* Helper function that allocates a new node with the given data and NULL left and right pointers. */ struct TNode* newNode( int data) { struct TNode* node = ( struct TNode*) malloc ( sizeof ( struct TNode)); node->data = data; node->left = NULL; node->right = NULL; return node; } /* A utility function to print preorder traversal of BST */ void preOrder( struct TNode* node) { if (node == NULL) return ; printf ( "%d " , node->data); preOrder(node->left); preOrder(node->right); } /* Driver program to test above functions */ int main() { int arr[] = {1, 2, 3, 4, 5, 6, 7}; int n = sizeof (arr)/ sizeof (arr[0]); /* Convert List to BST */ struct TNode *root = sortedArrayToBST(arr, 0, n-1); printf ( "n PreOrder Traversal of constructed BST " ); preOrder(root); return 0; } |
JAVA
// Java program to print BST in given range // A binary tree node class Node { int data; Node left, right; Node( int d) { data = d; left = right = null ; } } class BinaryTree { static Node root; /* A function that constructs Balanced Binary Search Tree from a sorted array */ Node sortedArrayToBST( int arr[], int start, int end) { /* Base Case */ if (start > end) { return null ; } /* Get the middle element and make it root */ int mid = (start + end) / 2 ; Node node = new Node(arr[mid]); /* Recursively construct the left subtree and make it left child of root */ node.left = sortedArrayToBST(arr, start, mid - 1 ); /* Recursively construct the right subtree and make it right child of root */ node.right = sortedArrayToBST(arr, mid + 1 , end); return node; } /* A utility function to print preorder traversal of BST */ void preOrder(Node node) { if (node == null ) { return ; } System.out.print(node.data + " " ); preOrder(node.left); preOrder(node.right); } public static void main(String[] args) { BinaryTree tree = new BinaryTree(); int arr[] = new int []{ 1 , 2 , 3 , 4 , 5 , 6 , 7 }; int n = arr.length; root = tree.sortedArrayToBST(arr, 0 , n - 1 ); System.out.println( "Preorder traversal of constructed BST" ); tree.preOrder(root); } } // This code has been contributed by Mayank Jaiswal |
python
# Python code to convert a sorted array # to a balanced Binary Search Tree # binary tree node class Node: def __init__( self , d): self .data = d self .left = None self .right = None # function to convert sorted array to a # balanced BST # input : sorted array of integers # output: root node of balanced BST def sortedArrayToBST(arr): if not arr: return None # find middle mid = ( len (arr)) / 2 # make the middle element the root root = Node(arr[mid]) # left subtree of root has all # values <arr[mid] root.left = sortedArrayToBST(arr[:mid]) # right subtree of root has all # values >arr[mid] root.right = sortedArrayToBST(arr[mid + 1 :]) return root # A utility function to print the preorder # traversal of the BST def preOrder(node): if not node: return print node.data, preOrder(node.left) preOrder(node.right) # driver program to test above function """ Constructed balanced BST is 4 / 2 6 / / 1 3 5 7 """ arr = [ 1 , 2 , 3 , 4 , 5 , 6 , 7 ] root = sortedArrayToBST(arr) print "PreOrder Traversal of constructed BST " , preOrder(root) # This code is contributed by Ishita Tripathi |
C#
using System; // C# program to print BST in given range // A binary tree node public class Node { public int data; public Node left, right; public Node( int d) { data = d; left = right = null ; } } public class BinaryTree { public static Node root; /* A function that constructs Balanced Binary Search Tree from a sorted array */ public virtual Node sortedArrayToBST( int [] arr, int start, int end) { /* Base Case */ if (start > end) { return null ; } /* Get the middle element and make it root */ int mid = (start + end) / 2; Node node = new Node(arr[mid]); /* Recursively construct the left subtree and make it left child of root */ node.left = sortedArrayToBST(arr, start, mid - 1); /* Recursively construct the right subtree and make it right child of root */ node.right = sortedArrayToBST(arr, mid + 1, end); return node; } /* A utility function to print preorder traversal of BST */ public virtual void preOrder(Node node) { if (node == null ) { return ; } Console.Write(node.data + " " ); preOrder(node.left); preOrder(node.right); } public static void Main( string [] args) { BinaryTree tree = new BinaryTree(); int [] arr = new int []{1, 2, 3, 4, 5, 6, 7}; int n = arr.Length; root = tree.sortedArrayToBST(arr, 0, n - 1); Console.WriteLine( "Preorder traversal of constructed BST" ); tree.preOrder(root); } } // This code is contributed by Shrikant13 |
Javascript
<script> // JavaScript program to print BST in given range // A binary tree node class Node { constructor(d) { this .data = d; this .left = null ; this .right = null ; } } var root = null ; /* A function that constructs Balanced Binary Search Tree from a sorted array */ function sortedArrayToBST(arr, start, end) { /* Base Case */ if (start > end) { return null ; } /* Get the middle element and make it root */ var mid = parseInt((start + end) / 2); var node = new Node(arr[mid]); /* Recursively construct the left subtree and make it left child of root */ node.left = sortedArrayToBST(arr, start, mid - 1); /* Recursively construct the right subtree and make it right child of root */ node.right = sortedArrayToBST(arr, mid + 1, end); return node; } /* A utility function to print preorder traversal of BST */ function preOrder(node) { if (node == null ) { return ; } document.write(node.data + " " ); preOrder(node.left); preOrder(node.right); } var arr = [1, 2, 3, 4, 5, 6, 7]; var n = arr.length; root = sortedArrayToBST(arr, 0, n - 1); document.write( "Preorder traversal of constructed BST<br>" ); preOrder(root); </script> |
输出：
Preorder traversal of constructed BST4 2 1 3 6 5 7
Tree representation of above output: 4 2 61 3 5 7
时间复杂性： O（n） 以下是SorterDarrayTobst（）的递归关系。
T(n) = 2T(n/2) + C T(n) --> Time taken for an array of size n C --> Constant (Finding middle of array and linking root to left and right subtrees take constant time)
上述复发可通过以下方法解决： 主定理 就像案例1中的情况一样。
如果您发现任何不正确的地方，或者您想分享有关上述主题的更多信息，请写下评论。
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