Maximum Binary Tree

Problem

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

The root is the maximum number in the array. The left subtree is the maximum tree constructed from left part subarray divided by the maximum number. 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].

Solution

Recursive Solution using two pointers: O(n^2) time and O(n) space

public class Solution {

    public TreeNode constructMaximumBinaryTree(int[] nums) {
        return helper(nums, 0, nums.length - 1);
    }

    private TreeNode helper(int[] nums, int left, int right) {
        if (left > right) return null;
        int maxIndex = left;
        for (int i = left + 1; i <= right; i++) {
            if (nums[i] > nums[maxIndex]) maxIndex = i;
        }
        TreeNode head = new TreeNode(nums[maxIndex]);
        head.left = helper(nums, left, maxIndex - 1);
        head.right = helper(nums, maxIndex + 1, right);
        return head;
    }
}

Analysis

We use a helper(int[] nums, int left, int right) to solve this problem easily
If left > right we know there is no valid indices and return null
We find the maxIndex from the given left to right
Then we create a new TreeNode with the value of maxIndex
We set it node.left and node.right by recursively calling the help() method
The left parameter will be the left to maxIndex - 1, and right will be maxIndex + 1 to right
Notice that the time complexity is O(n^2) because each recursive call costs O(n) time