Leetcode #872: Leaf-Similar Trees
In this guide, we solve Leetcode #872 Leaf-Similar Trees 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
Consider all the leaves of a binary tree, from left to right order, the values of those leaves form a leaf value sequence. For example, in the given tree above, the leaf value sequence is (6, 7, 4, 9, 8).
Quick Facts
- Difficulty: Easy
- Premium: No
- Tags: Tree, Depth-First Search, Binary Tree
Intuition
We need to explore a structure deeply before backing up, which suits DFS.
DFS keeps local context on the call stack and is easy to implement recursively.
Approach
Define a recursive DFS that carries the necessary state.
Combine child results as the recursion unwinds.
Steps:
- Define a recursive DFS with state.
- Visit children and combine results.
- Return the final aggregation.
Example
Input: root1 = [3,5,1,6,2,9,8,null,null,7,4], root2 = [3,5,1,6,7,4,2,null,null,null,null,null,null,9,8]
Output: true
Python Solution
# 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 leafSimilar(self, root1: Optional[TreeNode], root2: Optional[TreeNode]) -> bool:
def dfs(root: Optional[TreeNode], nums: List[int]) -> None:
if root.left == root.right:
nums.append(root.val)
return
if root.left:
dfs(root.left, nums)
if root.right:
dfs(root.right, nums)
l1, l2 = [], []
dfs(root1, l1)
dfs(root2, l2)
return l1 == l2
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.