【题目】
给出的升序排序的数组,个数必为奇数,要求形成二叉搜索(平衡)树。
【思路】
辅助函数fun,[0,len]=>[0,mid-1]+[mid+1,len]。
当left>right,返回null。
public TreeNode fun(int[] nums,int left,int right) { int mid=(right+left)/2; TreeNode tmp=new TreeNode(nums[mid]); if(right<left) return null; tmp.left=fun(nums,left,mid-1); tmp.right=fun(nums,mid+1,right); return tmp; }
【AC代码】
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public TreeNode sortedArrayToBST(int[] nums) { int len=nums.length; if(nums.length==0){ return null; } TreeNode root=fun(nums,0,len-1); return root; } public TreeNode fun(int[] nums,int left,int right) { int mid=(right+left)/2; TreeNode tmp=new TreeNode(nums[mid]); if(right<left) return null; tmp.left=fun(nums,left,mid-1); tmp.right=fun(nums,mid+1,right); return tmp; } }