Java递归加上二分搜索将有序数组转化为平衡二叉树

class Solution {
   public TreeNode sortedArrayToBST(int[] nums) {
		return ToBST(nums,0,nums.length-1);
	}
	public static TreeNode ToBST(int nums[],int left,int right){
		if(left>right)return null;//定义的二分区间为[left,right],无法进行继续递归,直接退出
		int mid = (int)(left+right)/2;//二分中值
		TreeNode root = new TreeNode(nums[mid]);
		root.left = ToBST(nums,left,mid-1);//注意mid-1 对左半部分进行递归
		root.right = ToBST(nums,mid+1,right);//注意mid+1 对右半部分进行递归
		return root;
	}
}

  

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转载自www.cnblogs.com/z2529827226/p/11746414.html