给定一个排序数组。编写一个使用数组元素创建平衡二进制搜索树的函数。
例子：
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
在上一篇文章中，我们讨论了从排序的链表构造BST。从O(n)时间中的排序数组构造起来比较简单，因为我们可以在O(1)时间中获得中间元素。以下是一个简单的算法，在该算法中，我们首先找到列表的中间节点，并将其设为要构建的树的根。
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.
以下是上述算法的实现。创建平衡BST的主要代码突出显示。
C++
// C++ program to print BST in given range
#include
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 \n";
preOrder(root);
return 0;
}
// This code is contributed by rathbhupendra C
#include
#include
/* 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 JaiswalPython
# 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.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输出：
Preorder traversal of constructed BST
4 2 1 3 6 5 7 Tree representation of above output:
4
2 6
1 3 5 7
时间复杂度： O(n)
以下是sortedArrayToBST()的重复关系。
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)
如果情况1发生上述情况，可以使用Master定理解决。