Convert Sorted Array to Binary Search Tree
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Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
这一题因为数组是已经排序好的。所以主要考虑点是在于:
1、作为平衡树,左边的结点的都要比右边的结点小,同时,左边的结点的个数和右边的结点个数之差小于等于1.所以,可以默认为 数组的中位数就是root,小于root的是root的左子树结点,大于root的是root的右子树结点。
接下来就是基本的找中位数的过程。
代码如下:
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */public class Solution { public TreeNode sortedArrayToBST(int[] nums) { if(nums.length==0) return null; return sortedArrayToBST(nums,0,nums.length-1); } private TreeNode sortedArrayToBST(int[] nums, int start, int end) { if(start>end) return null; int mid = (start+end)/2; TreeNode root = new TreeNode(nums[mid]); root.left = sortedArrayToBST(nums,start,mid-1); root.right = sortedArrayToBST(nums,mid+1,end); return root; }}
注意点如下:
一开始写的时候,初始赋值是nums.length-1,而不是 nums.length;
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- Convert Sorted Array to Binary Search Tree
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