Leetcode #617: Merge Two Binary Trees
In this guide, we solve Leetcode #617 Merge Two Binary 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
You are given two binary trees root1 and root2. Imagine that when you put one of them to cover the other, some nodes of the two trees are overlapped while the others are not.
Quick Facts
- Difficulty: Easy
- Premium: No
- Tags: Tree, Depth-First Search, Breadth-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 = [1,3,2,5], root2 = [2,1,3,null,4,null,7]
Output: [3,4,5,5,4,null,7]
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 mergeTrees(
self, root1: Optional[TreeNode], root2: Optional[TreeNode]
) -> Optional[TreeNode]:
if root1 is None:
return root2
if root2 is None:
return root1
node = TreeNode(root1.val + root2.val)
node.left = self.mergeTrees(root1.left, root2.left)
node.right = self.mergeTrees(root1.right, root2.right)
return node
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.