Leetcode 654. Maximum Binary Tree 最大二叉树 解题报告

就是给了一堆不重复的数组，然后挑选一个最大的作为root，然后root左边的和右边的有分别按照这个方法进行构建。
所以我做的方法也很简单咯，那么就每次找到这一段里最大的做root，然后递归左右两边。。简单粗暴的解法。。至于再高深的求各位大佬指教，我就不懂了

哦，对了，说起来倒是有个优化策略，可以设置上界，这样就不用每次都搜索了，喵喵

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].

解题代码 python

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def constructMaximumBinaryTree(self, nums):
        """
        :type nums: List[int]
        :rtype: TreeNode
        """
        def helper(start, end, upper_bound):
            if start > end:
                return None
            max_i = start
            p = start + 1
            while ums[max_i] < upper_bound & p <= end:
                if nums[p] > nums[max_i]:
                    max_i = p
                p += 1
            node = TreeNode(nums[max_i])
            node.left = helper(start, max_i - 1, nums[max_i] - 1)
            node.right = helper(max_i + 1, end, nums[max_i] - 1)
            return node
        return helper(0,len(nums) -1, 9999999)