Leetcode #1719: Number Of Ways To Reconstruct A Tree
In this guide, we solve Leetcode #1719 Number Of Ways To Reconstruct A 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
You are given an array pairs, where pairs[i] = [xi, yi], and: There are no duplicates. xi < yi Let ways be the number of rooted trees that satisfy the following conditions: The tree consists of nodes whose values appeared in pairs.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Tree, Graph
Intuition
The data forms a graph, so we should explore nodes and edges systematically.
A traversal ensures we visit each node once while maintaining the needed state.
Approach
Build an adjacency list and traverse with BFS or DFS.
Aggregate results as you visit nodes.
Steps:
- Build the graph.
- Traverse with BFS/DFS.
- Accumulate the required output.
Example
Input: pairs = [[1,2],[2,3]]
Output: 1
Explanation: There is exactly one valid rooted tree, which is shown in the above figure.
Python Solution
class Solution:
def checkWays(self, pairs: List[List[int]]) -> int:
g = [[False] * 510 for _ in range(510)]
v = defaultdict(list)
for x, y in pairs:
g[x][y] = g[y][x] = True
v[x].append(y)
v[y].append(x)
nodes = []
for i in range(510):
if v[i]:
nodes.append(i)
g[i][i] = True
nodes.sort(key=lambda x: len(v[x]))
equal = False
root = 0
for i, x in enumerate(nodes):
j = i + 1
while j < len(nodes) and not g[x][nodes[j]]:
j += 1
if j < len(nodes):
y = nodes[j]
if len(v[x]) == len(v[y]):
equal = True
for z in v[x]:
if not g[y][z]:
return 0
else:
root += 1
if root > 1:
return 0
return 2 if equal else 1
Complexity
The time complexity is O(V+E). The space complexity is O(V).
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.