Leetcode654. 最大二叉树 Maximum Binary Tree - Python 递归法

思路：

1.找出nums中最大值及其索引；

2.将最大值赋值给root.val；

3.将nums用索引分成左右两部分，分别递归传入递归函数；

注意：

4.root.left 和 root.right 分别接着左右生成好的子树

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def constructMaximumBinaryTree(self, nums: List[int]) -> Optional[TreeNode]:

        def constructTree(num: List[int]) -> Optional[TreeNode]:
            if num == []:
                return
            
            max, index = 0, 0
            for i, j in enumerate(num):
                if j > max:
                    max = j
                    index = i

            root = TreeNode(num[index])
            num_left = num[:index]
            num_right = num[index + 1 : ]
            root.left = constructTree(num_left)
            root.right = constructTree(num_right)
            return root
        
        return constructTree(nums)

用此方法可以不用在递归函数里反复新建数组。节约时间和空间。

class Solution:
    """最大二叉树 递归法"""

    def constructMaximumBinaryTree(self, nums: List[int]) -> TreeNode:
        return self.traversal(nums, 0, len(nums))
    
    def traversal(self, nums: List[int], begin: int, end: int) -> TreeNode:
        # 列表长度为0时返回空节点
        if begin == end:
            return None
        
        # 找到最大的值和其对应的下标
        max_index = begin
        for i in range(begin, end):
            if nums[i] > nums[max_index]:
                max_index = i
        
        # 构建当前节点
        root = TreeNode(nums[max_index])
        
        # 递归构建左右子树
        root.left = self.traversal(nums, begin, max_index)
        root.right = self.traversal(nums, max_index + 1, end)
        
        return root