小黑下午第一场面试被鸽，一切遇见随缘，继续第二场的leetcode之旅:654. 最大二叉树

news2025/1/11 8:38:13

小黑代码

# 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 dfs(start, end):
            # 递归推出条件
            if start > end:
                return
            # 获取最大值和最大值对应索引
            index = -1
            max_ = -1
            for i in range(start, end+1):
                if nums[i] > max_:
                    max_ = nums[i]
                    index = i
            node = TreeNode(val=max_)
            node.left = dfs(start, index-1)
            node.right = dfs(index+1, end)
            return node
        return dfs(0, len(nums)-1)

在这里插入图片描述

单调栈法

# 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]:
        n = len(nums)
        # 找到左右的“一大”
        left = [-1] * n
        right = [-1] * n
        nodes = []
        # 初始化单调栈
        q = []
        for i in range(n):
            # 将该结点插入数组
            node = TreeNode(val=nums[i])
            nodes.append(node)
            # 满足单调栈出栈条件 
            while q and nums[q[-1]] < nums[i]:
                top = q.pop()
                right[top] = i
            # 栈非空的话，赋予该元素左“一大”
            if q:
                left[i] = q[-1]
            q.append(i)
        # 通过左右"一大"进行构建二叉树
        print(right,left)
        root = None
        for i in range(n):
            # 根结点
            if left[i] == right[i] == -1:
                root = nodes[i]
            # 向右一大看齐
            elif left[i] == -1 or (right[i] != -1 and nums[left[i]] > nums[right[i]]):
                nodes[right[i]].left = nodes[i]
            else:
                nodes[left[i]].right = nodes[i]
        return root

在这里插入图片描述

优先栈改进

# 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]:
        nodes = []
        n = len(nums)
        # 初始化优先栈
        q = []
        for i in range(n):
            node = TreeNode(val=nums[i])
            # 优先栈工作
            while q and nums[i] > nums[q[-1]]:
                node.left = nodes[q[-1]]
                q.pop()
            if q:
                nodes[q[-1]].right = node
            q.append(i)
            nodes.append(node)
        return nodes[q[0]]