654. Maximum Binary Tree

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].

class Solution {
    public TreeNode constructMaximumBinaryTree(int[] nums) {
        return helper1(nums, 0, nums.length - 1);
    }
    
    public TreeNode helper1(int[] nums, int left, int right) {
        if(left > right) return null;
        int maxind = helper2(nums, left, right);
        TreeNode root = new TreeNode(nums[maxind]);
        root.left = helper1(nums, left, maxind - 1);
        root.right = helper1(nums, maxind + 1, right);
        return root;
    }
    public int helper2(int[] nums, int left, int right) {
        int cur = left;
        for(int i = left; i <= right; i++) {
            if(nums[i] > nums[cur]) cur = i;
        }
        return cur;
    }
}

首先找到maximum的index，然后把当前maximum作为root，left就递归调用nums，left，maxind - 1，right调用nums，maxind+1， right。

找maximum的index就直接找。

标签：Binary,right,nums,int,Tree,maximum,654,root,left
来源： https://www.cnblogs.com/wentiliangkaihua/p/13783531.html