Leetcode #1130: Minimum Cost Tree From Leaf Values
In this guide, we solve Leetcode #1130 Minimum Cost Tree From Leaf Values 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 an array arr of positive integers, consider all binary trees such that: Each node has either 0 or 2 children; The values of arr correspond to the values of each leaf in an in-order traversal of the tree. The value of each non-leaf node is equal to the product of the largest leaf value in its left and right subtree, respectively.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Stack, Greedy, Array, Dynamic Programming, Monotonic Stack
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: arr = [6,2,4]
Output: 32
Explanation: There are two possible trees shown.
The first has a non-leaf node sum 36, and the second has non-leaf node sum 32.
Python Solution
class Solution:
def mctFromLeafValues(self, arr: List[int]) -> int:
def dfs(i: int, j: int) -> Tuple:
if i == j:
return 0, arr[i]
s, mx = inf, -1
for k in range(i, j):
s1, mx1 = dfs(i, k)
s2, mx2 = dfs(k + 1, j)
t = s1 + s2 + mx1 * mx2
if s > t:
s = t
mx = max(mx1, mx2)
return s, mx
return dfs(0, len(arr) - 1)[0]
Complexity
The time complexity is , and the space complexity is . The space complexity is .
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.