[Leetcode] 0572. Subtree of Another Tree

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内容目录

Easy

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. The tree tree could also be considered as a subtree of itself.

Example 1:

Input: root = [3,4,5,1,2], subRoot = [4,1,2]
Output: true

Example 2:

Input: root = [3,4,5,1,2,null,null,null,null,0], subRoot = [4,1,2]
Output: false

Constraints:

The number of nodes in the root tree is in the range [1, 2000].
The number of nodes in the subRoot tree is in the range [1, 1000].
-10 ⁴ <= root.val <= 10 ⁴
-10 ⁴ <= subRoot.val <= 10 ⁴

Python

				
					# time complexity: O(m*n) # space complexity: O(m+n) from typing import Optional class TreeNode: def __init__(self, val=0, left=None, right=None): self.val = val self.left = left self.right = right class Solution: def isSameTree(self, root1: Optional[TreeNode], root2: Optional[TreeNode]) -> bool: if not root1 and not root2: return True if not root1 or not root2: return False if root1.val != root2.val: return False return self.isSameTree(root1.left, root2.left) and self.isSameTree(root1.right, root2.right) def isSubtree(self, root: Optional[TreeNode], subRoot: Optional[TreeNode]) -> bool: def dfs(node): if node is None: return False elif self.isSameTree(node, subRoot): return True return dfs(node.left) or dfs(node.right) return dfs(root) root = TreeNode(3) root.left = TreeNode(4) root.left.left = TreeNode(1) root.left.right = TreeNode(2) root.right = TreeNode(5) subRoot = TreeNode(4) subRoot.left = TreeNode(1) subRoot.right = TreeNode(2) print(Solution().isSubtree(root, subRoot))

Easy

Python

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