标签：[] span class sort bin creat middle this geeks
Given a sorted array. Write a program that creates a Balanced Binary Search Tree using array elements.
If there are n elements in array, then floor(n/2)‘th element should be chosen as root and same should be followed recursively.
Solution.
1. get the middle element and create root node N.
2. recursively create a balanced BST from the left half of the array and set its root as N‘s left node.
3. recursively create a balanced BST from the right half of the array and set its root as N‘s right node.
T(n) = 2 * T(n/2) + O(1), so the time complexity is O(n).
1 class TreeNode { 2 TreeNode left; 3 TreeNode right; 4 int val; 5 TreeNode(int val){ 6 this.left = null; 7 this.right = null; 8 this.val = val; 9 } 10 } 11 public class Solution { 12 public TreeNode sortedArrayToBalancedBST(int[] a) { 13 if(a == null || a.length == 0){ 14 return null; 15 } 16 return toBSTRecursive(a, 0, a.length - 1); 17 } 18 private TreeNode toBSTRecursive(int[] a, int start, int end){ 19 if(start > end){ 20 return null; 21 } 22 int mid = start + (end - start) / 2; 23 TreeNode node = new TreeNode(a[mid]); 24 node.left = toBSTRecursive(a, start, mid - 1); 25 node.right = toBSTRecursive(a, mid + 1, end); 26 return node; 27 } 28 }
[GeeksForGeeks] Sorted array to balanced BST
标签：[] span class sort bin creat middle this geeks
原文地址：http://www.cnblogs.com/lz87/p/7282826.html
