Leetcode #427: Construct Quad Tree
In this guide, we solve Leetcode #427 Construct Quad Tree in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Given a n * n matrix grid of 0's and 1's only. We want to represent grid with a Quad-Tree.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Tree, Array, Divide and Conquer, Matrix
Intuition
The input is a tree, so recursive decomposition is a natural fit.
We can compute the answer by combining results from left and right subtrees.
Approach
Use DFS and pass the required state through recursive calls.
Combine child results to compute the answer for each node.
Steps:
- Pick traversal order.
- Recurse with state.
- Combine results from children.
Example
class Node {
public boolean val;
public boolean isLeaf;
public Node topLeft;
public Node topRight;
public Node bottomLeft;
public Node bottomRight;
}
Python Solution
"""
# 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':
def dfs(a, b, c, d):
zero = one = 0
for i in range(a, c + 1):
for j in range(b, d + 1):
if grid[i][j] == 0:
zero = 1
else:
one = 1
isLeaf = zero + one == 1
val = isLeaf and one
if isLeaf:
return Node(grid[a][b], True)
topLeft = dfs(a, b, (a + c) // 2, (b + d) // 2)
topRight = dfs(a, (b + d) // 2 + 1, (a + c) // 2, d)
bottomLeft = dfs((a + c) // 2 + 1, b, c, (b + d) // 2)
bottomRight = dfs((a + c) // 2 + 1, (b + d) // 2 + 1, c, d)
return Node(val, isLeaf, topLeft, topRight, bottomLeft, bottomRight)
return dfs(0, 0, len(grid) - 1, len(grid[0]) - 1)
Complexity
The time complexity is O(n). The space complexity is O(h).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.