Update 2.树.md

Author: scu-zzy

2.树.md

@@ -1443,19 +1443,18 @@ BFS，使用队列，用一个标志位来决定是否需要翻转。
 高度平衡意味着每次必须选择中间数字作为根节点。
 
 	class Solution {
+	    public TreeNode sortedArrayToBST(int[] nums) {
+	        return buildBST(nums, 0, nums.length-1);
+	    }
 
-	    TreeNode helper(int left, int right, int[] nums){
-	        if(left > right) return null;
-	        int mid = (left + right) / 2;
+	    private TreeNode buildBST(int[] nums, int start, int end){
+	        if(start > end) return null;
+	        int mid = (start + end)/2;
 	        TreeNode root = new TreeNode(nums[mid]);
-	        root.left = helper(left, mid - 1, nums);
-	        root.right = helper(mid+1, right, nums);
+	        root.left = buildBST(nums,start,mid-1);
+	        root.right = buildBST(nums,mid+1,end);
 	        return root;
 	    }
-	
-	    public TreeNode sortedArrayToBST(int[] nums) {
-	        return helper(0,nums.length-1,nums);
-	    }
 	}
 
 ## 7. 根据有序链表构造平衡的二叉查找树