Leetcode #1644: Lowest Common Ancestor of a Binary Tree II
In this guide, we solve Leetcode #1644 Lowest Common Ancestor of a Binary Tree II 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 the root of a binary tree, return the lowest common ancestor (LCA) of two given nodes, p and q. If either node p or q does not exist in the tree, return null.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- 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: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
Output: 3
Explanation: The LCA of nodes 5 and 1 is 3.
Python Solution
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def lowestCommonAncestor(
self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode'
) -> 'TreeNode':
def dfs(root, p, q):
if root is None:
return False
l = dfs(root.left, p, q)
r = dfs(root.right, p, q)
nonlocal ans
if l and r:
ans = root
if (l or r) and (root.val == p.val or root.val == q.val):
ans = root
return l or r or root.val == p.val or root.val == q.val
ans = None
dfs(root, p, q)
return ans
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.