[leetcode] 427. Construct Quad Tree @ python_leetcode427 python-优快云博客

本文链接：https://blog.youkuaiyun.com/danspace1/article/details/87901869

本文详细解析了LeetCode上构建四叉树的问题，通过递归算法实现对二维网格的分解，判断每个子网格是否为叶子节点，并进一步构建四叉树结构。关键步骤包括：求网格总和、判断叶子节点、划分子网格并递归调用。

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原题

https://leetcode.com/problems/construct-quad-tree/

解法

递归. Base case是grid为空, 此时返回None.
先求grid的和, 以确定grid是否为叶子节点. 如果grid的所有值都为0 或1, 那么grid所组成的是叶子节点, 值分别为False或者True. 否则grid不是叶子节点, 它的值为True(象限四分数中只要有一个子树的值为True, 那么它的值为True), 然后构造subgrid, 递归求解.

代码

"""
# Definition for a QuadTree node.
class Node:
    def __init__(self, val, isLeaf, topLeft, topRight, bottomLeft, bottomRight):
        self.val = val
        self.isLeaf = isLeaf
        self.topLeft = topLeft
        self.topRight = topRight
        self.bottomLeft = bottomLeft
        self.bottomRight = bottomRight
"""
class Solution:
    def construct(self, grid: List[List[int]]) -> 'Node':
        # base case
        if not grid: return None
        
        total = sum([sum(row) for row in grid])
        l = len(grid)
        
        if total == 0:
            root =  Node(False, True, None, None, None, None)
            return root
            
        elif total == l**2:
            root =  Node(True, True, None, None, None, None)
            return root
            
        else:
            mid = l//2
            topLeft = [[grid[i][j] for j in range(mid)] for i in range(mid)]
            topRight = [[grid[i][j] for j in range(mid, l)] for i in range(mid)]
            bottomLeft = [[grid[i][j] for j in range(mid)] for i in range(mid, l)]
            bottomRight = [[grid[i][j] for j in range(mid, l)] for i in range(mid, l)]
            root = Node(True, False, self.construct(topLeft), self.construct(topRight), self.construct(bottomLeft), self.construct(bottomRight))
            
            return root