# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
# https://leetcode.com/problems/subtree-of-another-tree/solution/
def isSubtree(self, s, t):
"""
:type s: TreeNode
:type t: TreeNode
:rtype: bool
"""
s_res = self.preorder(s, True)
t_res = self.preorder(t, True)
return t_res in s_res
def preorder(self, root, isLeft):
if root is None:
if isLeft:
return "lnull"
else:
return "rnull"
return "#" + str(root.val) + " " + self.preorder(root.left, True) + " " + self.preorder(root.right, False)
# def isSubtree(self, s, t):
# return self.traverse(s, t)
# def equals(self, x, y):
# if x is None and y is None:
# return True
# if x is None or y is None:
# return False
# return x.val == y.val and self.equals(x.left, y.left) and self.equals(x.right, y.right)
# def traverse(self, s, t):
# return s is not None and (self.equals(s, t) or self.traverse(s.left, t) or self.traverse(s.right, t))
Download Subtree of Another Tree.pyLeetcode 572 Subtree of Another Tree problem solution in python3 with explanation. This is the best place to expand your knowledge and get prepared for your next interview.
self.val = x
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