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Leetcode #572: Subtree of Another Tree

In this guide, we solve Leetcode #572 Subtree of Another Tree 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 roots of two binary trees root and subRoot, return true if there is a subtree of root with the same structure and node values of subRoot and false otherwise. A subtree of a binary tree tree is a tree that consists of a node in tree and all of this node's descendants.

Quick Facts

  • Difficulty: Easy
  • Premium: No
  • Tags: Tree, Depth-First Search, Binary Tree, String Matching, Hash Function

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 = [3,4,5,1,2], subRoot = [4,1,2] 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 isSubtree(self, root: Optional[TreeNode], subRoot: Optional[TreeNode]) -> bool: def same(p: Optional[TreeNode], q: Optional[TreeNode]) -> bool: if p is None or q is None: return p is q return p.val == q.val and same(p.left, q.left) and same(p.right, q.right) if root is None: return False return ( same(root, subRoot) or self.isSubtree(root.left, subRoot) or self.isSubtree(root.right, subRoot) )

Complexity

The time complexity is O(n×m)O(n \times m)O(n×m), and the space complexity is O(n)O(n)O(n). The space complexity is O(n)O(n)O(n).

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.


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