思路： 构建二叉树 一定想到是前序 ： 中左右

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode constructMaximumBinaryTree(int[] nums) {
        //构建二叉树一定想到是前序 ： 中左右
        //数组最少有一个时返回
        if(nums.length == 1) {
            return new TreeNode(nums[0]);
        }
        int max = 0;
        int index =0;
        //1、中，找到数组中最大的值，并记录下标
        for(int i=0; i<nums.length;i++){
            if(max < nums[i]){
                max = nums[i];
                index = i;
            }
        }
        //当前最大值作为头节点
        TreeNode node = new TreeNode(max);
        //2、左，数组元素至少1个：index > 0
        if(index > 0) {
            //截取左数组放入继续处理，下标左闭右开
            int[] newArray = Arrays.copyOfRange(nums, 0, index);
            node.left = constructMaximumBinaryTree(newArray);
        }
        //3、右，数组元素至少1个：index < nums.length-1
        if(index < nums.length-1 ) {
            //截取右数组放入继续处理，下标左闭右开
            int[] newArray2 = Arrays.copyOfRange(nums, index+1, nums.length );
            node.right = constructMaximumBinaryTree(newArray2);
        }
        return node;

    }
}