Leetcode #1676: Lowest Common Ancestor of a Binary Tree IV

In this guide, we solve Leetcode #1676 Lowest Common Ancestor of a Binary Tree IV 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.

Leetcode

Problem Statement

Given the root of a binary tree and an array of TreeNode objects nodes, return the lowest common ancestor (LCA) of all the nodes in nodes. All the nodes will exist in the tree, and all values of the tree's nodes are unique.

Quick Facts

Difficulty: Medium
Premium: Yes
Tags: Tree, Depth-First Search, Hash Table, Binary Tree

Intuition

Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.

By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.

Approach

Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.

This keeps the solution linear while remaining easy to explain in an interview setting.

Steps:

Initialize a hash map for seen items or counts.
Iterate through the input, querying/updating the map.
Return the first valid result or the final computed value.

Example

Input: root = [3,5,1,6,2,0,8,null,null,7,4], nodes = [4,7]
Output: 2
Explanation: The lowest common ancestor of nodes 4 and 7 is node 2.

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', nodes: 'List[TreeNode]'
    ) -> 'TreeNode':
        def dfs(root):
            if root is None or root.val in s:
                return root
            left, right = dfs(root.left), dfs(root.right)
            if left and right:
                return root
            return left or right

        s = {node.val for node in nodes}
        return dfs(root)

Complexity

The time complexity is $O(n + m)$ , and the space complexity is $O(n + m)$ . The space complexity is $O(n + m)$ .

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.

Approach

Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.

This keeps the solution linear while remaining easy to explain in an interview setting.

Steps:

Initialize a hash map for seen items or counts.

Iterate through the input, querying/updating the map.

Return the first valid result or the final computed value.

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', nodes: 'List[TreeNode]'
    ) -> 'TreeNode':
        def dfs(root):
            if root is None or root.val in s:
                return root
            left, right = dfs(root.left), dfs(root.right)
            if left and right:
                return root
            return left or right

        s = {node.val for node in nodes}
        return dfs(root)

Leetcode #1676: Lowest Common Ancestor of a Binary Tree IV

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview

Leetcode #1676: Lowest Common Ancestor of a Binary Tree IV

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview