LeetCode-Maximum Binary Tree

Description：
Given an integer array with no duplicates. A maximum tree building on this array is defined as follow:

1. The root is the maximum number in the array.
2. The left subtree is the maximum tree constructed from left part subarray divided by the maximum number.
3. The right subtree is the maximum tree constructed from right part subarray divided by the maximum number.

Construct the maximum tree by the given array and output the root node of this tree.

Example 1:

Input: [3,2,1,6,0,5]
Output: return the tree root node representing the following tree:

      6
    /   \
   3     5
    \    / 
     2  0   
       \
        1

Note:

• The size of the given array will be in the range [1,1000].

题意：给定一个一维数组，里面的元素代表树的节点；现在要求构造一颗maximum tree；具体要求如下

• 根节点的值为数组中最大的那个数
• 左子树的根节点的值为最大数所在数组位置左半部分数组中的最大值
• 右子树的根节点的值为最大数所在数组位置右半部分数组中的最大值

解法：我们采用递归的方法来解决这道题目；首先从数组中找到最大值最为根节点的值，然后依次找左子树节点的值和右子树节点的值；我们需要用一个数组来标记访问过的元素，在判断终止条件时有下面的两种情况（其中，st表示要遍历的数组的开始坐标，ed表示要遍历的数组的结束坐标，visited数组表示访问过的数组位置）

• ed > st 表示没有左节点或者右节点可以选择
• ed == st && visited[st] 仅有一个数可供选择，但是已经被访问过

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode constructMaximumBinaryTree(int[] nums) {
        TreeNode root = null;
        boolean[] visited = new boolean[nums.length];
        root = getRoot(root, nums, visited, 0, nums.length - 1);
        return root;
    }
    
    private TreeNode getRoot(TreeNode root, int[] nums, boolean[] visited, int st, int ed) {
        if ((st == ed && visited[st]) || st > ed) {
            return null;
        }
        int index = st;
        root = new TreeNode(Integer.MIN_VALUE);
        for (int i = st; i <= ed; i++) {
            if (!visited[i] && nums[i] > root.val) {
                root.val = nums[i];
                index = i;
            }
        }
        visited[index] = true;
        root.left = getRoot(root.left, nums, visited, st, index - 1);
        root.right = getRoot(root.right, nums, visited, index + 1, ed);
        return root;
    }
}