654 Maximum Binary Tree

654. Maximum Binary Tree

1. Question

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:

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

2. Implementation

思路：采用Divide and Conquer的思想，在每个区间里找到最大值作为该区间的root，然后根据root在区间的index，将区间一分为二，分别找出root的左subtree和右subtree的root

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

    public TreeNode constructMaxBinaryTree(int[] nums, int start, int end) {
        if (start > end) {
            return null;
        }

        int index = getIndexOfMax(nums, start, end);
        TreeNode curNode = new TreeNode(nums[index]);

        curNode.left = constructMaxBinaryTree(nums, start, index - 1);
        curNode.right = constructMaxBinaryTree(nums, index + 1, end);
        return curNode;
    }

    public int getIndexOfMax(int[] nums, int start, int end) {
        int res = start;
        for (int i = start + 1; i <= end; i++) {
            if (nums[res] < nums[i]) {
                res = i;
            }
        }
        return res;
    }
}

3. Time & Space Complexity

时间复杂度是O(nlogn), 空间复杂度是O(logn)