题解 | #将升序数组转化为平衡二叉搜索树#

using System;
using System.Collections.Generic;

/*
public class TreeNode
{
    public int val;
    public TreeNode left;
    public TreeNode right;

    public TreeNode (int x)
    {
        val = x;
    }
}
*/

class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     *
     * @param nums int整型一维数组
     * @return TreeNode类
     */
    public TreeNode sortedArrayToBST (List<int> nums) {
        // write code here
        if (nums == null || nums.Count == 0)
            return null;
        if (nums.Count <= 1)
            return new TreeNode(nums[0]);
        return PartInitTree(nums, 0, nums.Count - 1);
    }

    public static TreeNode PartInitTree(List<int> nums, int nL, int nR) {
        if (nL > nR)
            return null;
        int nRootIndex = (int)Math.Ceiling((nR + nL) / 2.0);
        TreeNode treeNode = new TreeNode(nums[nRootIndex]);
        var t1 = PartInitTree(nums, nL, nRootIndex - 1);
        if (t1 == null)
            treeNode.left = t1;
        else {
            if (treeNode.val > t1.val)
                treeNode.left = t1;
            else
                treeNode.right = t1;
        }
        var t2 = PartInitTree(nums, nRootIndex + 1, nR);
        if (t2 == null)
            treeNode.right = t2;
        else {
            if (treeNode.val > t2.val)
                treeNode.left = t2;
            else
                treeNode.right = t2;
        }
        return treeNode;
    }
}