654. Maximum Binary Tree 最大二叉树

Posted 2020-10-02 Long Long Journey

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].

根据所给数组建立二叉树，最大的作为根节点，左侧部分作为左子树，右侧部分作为右子树

# 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
        """
        if not nums:
            return None
        def helper(node, nums):
            index = nums.index(max(nums))
            leftArr = nums[:index]
            rightArr = nums[index + 1:]
            if leftArr:
                node.left = TreeNode(max(leftArr))
                helper(node.left, leftArr)
            if rightArr:
                node.right = TreeNode(max(rightArr))
                helper(node.right, rightArr)
        root = TreeNode(max(nums))
        helper(root, nums)
        return root

来自为知笔记(Wiz)