分析
难度 易
来源
https://leetcode.com/problems/convert-sorted-array-to-binary-search-tree/
题目
Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.
Example:
Given the sorted array: [-10,-3,0,5,9],
One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:
0
/ \
-3 9
/ /
-10 5
解答 1 package LeetCode; 2 3 import java.util.LinkedList; 4 import java.util.Queue; 5 6 public class L108_ConvertSortedArray2BinarySearchTree { 7 public TreeNode makeBinaryTreeByArray(int[] nums,int low,int high){ 8 if(low<=high){ 9 int mid=(high+low)/2; 10 TreeNode root=new TreeNode(nums[mid]); 11 //System.out.println(root.val); 12 root.left=makeBinaryTreeByArray(nums,low,mid-1); 13 root.right=makeBinaryTreeByArray(nums,mid+1,high); 14 return root; 15 } 16 else 17 return null; 18 } 19 public TreeNode sortedArrayToBST(int[] nums) { 20 return makeBinaryTreeByArray(nums,0,nums.length-1); 21 } 22 public void levelOrderTraversal(TreeNode root){//广度优先搜索+分层 23 if(root==null){ 24 //System.out.println("empty tree"); 25 return; 26 } 27 Queue<TreeNode> queue=new LinkedList<TreeNode>(); 28 queue.add(root); 29 int curCount=1;//记录当前层节点数 30 int nextCount=0;//记录下一层节点数 31 int outCount=0;//记录出队列节点数 32 while(!queue.isEmpty()){ 33 TreeNode node=queue.remove(); 34 outCount++; 35 System.out.print(node.val+"\t"); 36 if(node.left!=null){ 37 queue.add(node.left); 38 nextCount++; 39 } 40 if(node.right!=null){ 41 queue.add(node.right); 42 nextCount++; 43 } 44 if(outCount==curCount)//当前层全部出队列 45 { 46 System.out.println(); 47 curCount=nextCount; 48 outCount=0; 49 } 50 } 51 System.out.println(""); 52 } 53 public static void main(String[] args){ 54 int[] nums={-10,-3,0,5,9}; 55 L108_ConvertSortedArray2BinarySearchTree l108=new L108_ConvertSortedArray2BinarySearchTree(); 56 TreeNode root=l108.sortedArrayToBST(nums); 57 l108.levelOrderTraversal(root); 58 } 59 }